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ELEMENTS 



PLANE TRIGONOMETRY 



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BY WILLIAM SMYTH, A. M. 

PROFESSOR OF MATHEMATICS IN BOWDOIN COLLEGE. 



b 




BOSTON. 

LILLY, WAIT, COL MAN, AND HOLDEN. 

1834. 



Entered, according to act of Congress, in the year 1834, by William 
Smyth, A. M. in the Clerk's Office of the District of Maine. 






-*•*!& 



I»RESS OF J. GRIFFIN, BRUNSWICK, ME. 



ADVERTISEMENT. 



In the following treatise the writer has endeavored to 
present the principles of plane trigonometry in a natural 
and connected order, to show the application of each 
principle as it is introduced, and in general to lead the 
student to feel the want of a new principle before pro- 
ceeding to its investigation. The work has been prepar- 
ed chiefly with reference to a college class. Regard has 
nevertheless been had to the wants of students in our 
public seminaries generally, in which trigonometry and 
its applications are taught. Several of the more recent 
treatises on trigonometry have been consulted in the 
course of the work ; the most assistance however has 
been derived from those of Lacroix and Bezout. 

The Geometry referred to is that of Legendre in the 
Cambridge Mathematics. 

Bowdoin College, Jan. 1834. 



TABLE OF CONTENTS. 



SECTION. PAGE. 

I. Preliminary Remarks 5 

II. Of the Solution of Triangles by Geometrical Construction . 6 

III. Of the Solution of Triangles by Arithmetical Calculation . . 12 

IV. Of the method of forming a Series of Triangles, having angles 

of every possible magnitude . t 15 

V. Of the Method of calculating the sides in the Series of Trian- 

gles constructed in the Quadrant of a Circle 17 

VI. Of the Solution of Triangles by the Table of Sines and Cosines 24 

VII. Of the second Method of constructing and calculating a Series 

of Triangles, having angles of every possible value . - 25 

VIII. Of the Solution of Triangles by the Table of Tangents and 

Secants ,..,...., 28 

IX. Of Cotangents and Cosecants ... ■ 30 

X. Miscellaneous Remarks on the Trigonometrical Tables . . 31 

XI. Of the Solution of oblique angled Triangles ...... 33 

XII. Mensuration of Heights and distances 40 

XIII. Of the Line of natural Sines, Tangents, &c 50 

XIV. Of the Line of logarithmic Sines, Tangents, &c 51 

XV. Trigonometrical Analysis 55 



ELEMENTS 



PIiAl¥E TRIGONOMETRY. 



SECTION I. 

Preliminary Remarks. 

1. The triangle ABC (fig. 1), bounded by right lines, 
and having in consequence all its parts situated in the same 
plane, is called a plane triangle. 

2. The plane triangle DEF (fig. 2), in which one of the 
sides DE forms with another EF a right angle, is called a 
right angled triangle. Any other triangle ABC (fig. 1), in 
which neither of the angles is a right angle, is called an 
oblique triangle. 

3. In the right angled triangle DEF (fig, 2), the part DF 
opposite the right angle, is called the hypothenuse ; the other 
two parts DE, EF are called the sides ; sometimes also the 
perpendicular and base. 

4. In a plane triangle there are six parts to be considered, 
viz, three angles and three sides. In order to calculate the 
rest certain of these parts must be given. 

5. Let the lines a and b (fig. 3) be two of the sides of a 
proposed triangle. These lines may evidently make with 

each other any angle whatever ; of consequence they may 

2 



6 PLANE TRIGONOMETRY. 

be sides in an indefinite number of triangles ABC, ABC', 
&c. The remaining parts of a triangle cannot therefore be 
determined, when two of the sides are the only parts, which 
are given. Again let one of the sides AB and an angle A 
(fig. 4) be two known parts in a proposed triangle ; it is 
evident, that these alone will not be sufficient to determine 
the rest ; for the side AB and the angle A may also belong 
to an indefinite number of triangles ABC, ABC, &c. 
Moreover if two of the angles, or indeed if all the angles, 
are the only parts, which are given, the remaining parts 
must still be indeterminate ; for in the triangle ABC (fig. 5,) 
if the lines B' C, B" C", &c. be drawn parallel to BC, 
an indefinite number of triangles may be formed, in each of 
which the angles will be respectively equal, while the sides 
are different in value. In order then, that a triangle maybe 
calculated, three at least of its parts should be given, one of which 
must be a side. 

6. That part of trigonometry, which relates to the calcula- 
tion of plane triangles, is called Plane Trigonometry. 

SECTION II. 

Of the solution of Triangles by Geometrical construction. 

7. In order to calculate a triangle geometrically, its dif- 
ferent parts must be measured ; among these parts there are 
two kinds of quantities, viz, angles and sides. 

8. Since the sides of plane triangles are right lines, their 
values are expressed in the usual measures of extension, 
as feet, yards, &c. To represent these measures for the 
purpose of calculating a triangle geometrically, any line 
AB (fig. 6) may be divided into a convenient number of 
equal parts ; one of these being considered as unity may be 
further divided into equal portions for fractional parts of 
n ity. A line divided in this manner is called a scale of 
equal parts . 



PLANE TRIGONOMETRY. 7 

In the example figure 6, the line AB is divided into 
eleven equal portions ; of these the extreme one on the left 
is divided into ten equal parts, the remainder being number- 
ed 1, 2, 3, &c. The principal divisions of this scale may be 
considered as feet, miles, &c; then the subdivisions will be 
tenths of a foot, mile, &c. The principal divisions may also 
be regarded as ten feet, ten miles, &c; in this case the sub- 
divisions will represent feet, miles, &c. 

A convenient construction for a scale of equal parts is 
represented figure 7. It consists, when intended for a de- 
cimal scale, of eleven lines drawn parallel to each other at 
equal distances, and divided into convenient portions by 
perpendicular lines. In the extreme division on the left, the 
upper and lower lines are divided into ten equal parts, and 
the subdividing points are connected by diagonal lines, that 
is, bylines drawn from the first point on the lower line to 
the second on the upper. Then by similar triangles ac : ab 
: : cd : 6o, that is, cd is one tenth of the subdivisions, or one 
hundredth of the primary divisions of the scale. In like 
manner e/may be shown to be two hundredths, and gh three 
hundredths of a\. 

Let it be required, for example, to take off the number 
255 from this scale. Considering the subdivisions as con- 
taining ten each, we extend the compasses from figure 2 of 
the principal divisions to 5 of the subdivisions for 250 ; then 
by opening the compasses to the corresponding extent on the 
fifth of the parallel lines, we shall obtain the length required. 

9. In order to estimate the magnitude of the angles, it 
would seem more natural to refer them to the right angle, as 
the unit of measure ; since, in this case, the quantities to be 
measured would be the same in kind with the quantity as- 
sumed as their measure. It is found, however, more con- 
venient in practice to measure angles by arcs of circles. In 



8 PLANE TRIGONOMETRY. 

the circle AC (fig. 8), in whatever ratio the angle DBC at 
the centre increases or diminishes, the arc CD, on which it 
stands, increases or diminishes in the same ratio (Geom, 
122); the arc CD may therefore be assumed, as the measure 
of this angle. In like manner any other concentric arc EF, 
intercepted by the sides of the angle DBC, may be used as 
its measure. The arc CD, indeed, considered as a magni- 
tude of a certain length, is manifestly greater than the arc 
EF. In the measure of angles, however, it is not the abso- 
lute value of the arcs, which we regard, but merely their 
ratio to an entire circumference. Since then the arc EF is 
to the circumference EG, as the arc CD to the circumference 
CA ; considered as parts of the entire circumferences, to 
which they respectively belong, the arcs CD and EF are 
equal in quantity. The measure of an angle is then the arc of a 
circle, having its centre in the vertex of the angle , and intercepted 
by the lines, which form its sides. In the measure of an angle it 
is therefore immaterial, what may be the size of the circle 
employed, provided it cut the sides of the angle. In com- 
paring different angles, however, the arcs, on which they 
are measured, should be described with equal radii. 

10. The entire circumference of every circle, whether 
great or small, is divided into 360 equal parts called degrees, 
each degree into 60 equal parts called minutes, each minute 
into 60 equal parts called seconds, each second into 60 equal 
parts called thirds, &c. The character used to denote de- 
grees is °, placed at the right hand of the unit figure and a 
little above it, that for minutes is ', that for seconds is ", 
that for thirds is '", so that 40° 30' 25" 10"' signifies 40 
degrees, 30 minutes, 25 seconds, and 10 thirds. This divis- 
ion, however, is entirely arbitrary, and any other may be 
assumed at pleasure. We may, for instance, after the man- 
ner of the French, divide the circle into 400 equal parts for 



PLANE TRIGONOMETRY. 9 

degrees, and each degree into 100 equal parts for minutes, 
with further subdivisions according to the principles of the 
decimal progression. 

11. We have seen art. 9, that the measure of any angle 
DEC (fig. 8) is the arc CD of a circle CA ; but the chord 
CD, being the distance in a right line between the extremi- 
ties of the arc CD, will determine the magnitude of this arc. 
If then the several chords for each degree, minute, &c, con- 
tained in the arc AB (fig. 9), equal to a quadrant or fourth 
part of a circle, be transferred to the line BD, the line BD, 
denominated in this case a line of chords, may with conven- 
ience be employed in the measurement of angles. 

Let it be required for example, to ascertain the magni- 
tude of the angle ABC (fig. 10) by the line of chords. With 
the extent from to 60 on the line of chords, we describe 
from B as a centre, an arc cutting the two lines, which in- 
clude the angle in the points m and n ; then since the chord 
of 60° is equal to radius (Geom. 271), this arc will have the 
same radius with that, to which the given scale of chords 
belongs ; consequently, the distance between the points of 
intersection m and w, applied to the line of chords, will give 
the measure of the angle- in degrees. 

Conversely, let it be required to make an angle ABC of a 
given magnitude, 30° for example. Having drawn a line 
BC, from the point B with a radius Brc, equal to the chord 
of 60° on the line of chords, describe an arcnm ; the extent 
from B to 30° on the line of chords applied from n on the 
arc nm will give the point, through which if the line BA be 
drawn, the angle ABC will be formed of the required mag- 
nitude. 

12. Let there now be given two of the sides of a triangle 
and the angle contained between them, to find, by construc- 
tion, the remaining sides and angles. From any convenient 

*2 



10 PLANE TRIGONOMETRY. 

scale of equal parts we take the line AB (fig. 1), equal to 
one of the given sides ; at the point A, one of the extremi- 
ties of this line, we make the angle A equal to the given an- 
gle, and draw through this the line AC equal to the other 
given side ; we then connect the points B and C by the line 
BC ; the triangle being thus constructed, the side BC, and 
the remaining angles may be measured. 

In general to calculate a triangle geometrically, we con- 
struct the triangle by means of the parts, which are given; 
the sides are laid clown from a scale of equal parts, and 
the angles from a line of chords 5 the required parts are 
then measured, the sides upon the same scale of equal parts, 
and the angles upon the same .line of chords, from which the 
given parts were laid down. 

The following examples exhibit the different cases, 
which occur in the solution of triangles. 

13. Prob. 1 . The three sides of a triangle being given, to 
find the angles. 

Ex. 1. In the triangle ABC (fig. 1), given the side AB 
39 rods, BC 45, and AC 38 rods ; to find the angles. 

Ex. 2. Given the three sides 53.5, 82.08, and 46.7; to 
find the angles. 

14. Prob. 2. Two sides and the included angle being given, 
to find the other side and angles. 

Ex. 1. In the triangle ABC (fig. 11), given the side AB 
60 rods, the side BC 100 rods, and the angle B 45° 30' ; re- 
quired the remaining parts. 

Ex. 2. Given two of the sides of a triangle equal to 345 
and 759 feet respectively, and the angle contained between 
them 53° 45', to find the remaining parts. 

15. Prob. 3. Two sides and an angle opposite one of them being 
given, to find the remaining side and the other two angles. 

Ex. 1. Given inthe triangle ABC (fig. 12) the side AB 



PLANE TRIGONOMETRY. U 

57.5 feet, the side BC 103 feet, and the angle A 99° 15'; re- 
quired the third side and the two other angles. 

Ex 2. Given in the triangle ABC (fig. 13) the side AB 72 
feet, the side AC 41.5 feet, and the angle B 30° ; required 
the remaining parts. 

16. In this last example, in which the side opposite the 
given angle is less, than the other given side, the solution is 
ambiguous ; for if from the point A, with a radius equal to 
AC, we describe the arc CC cutting the line BC in C, and 
draw AC, a second triangle ABC will be formed, which 
will also contain the three given parts. To determine then, 
which of these triangles is intended in the question proposed; 
we must know, in addition to the given parts, whether the 
angle C be acute or obtuse. With respect to this case, we 
may remark in passing, that the angles ACB, ACB are to- 
gether equal to two right angles. Indeed the angles ACB, 
ACC are together equal to two right angles, but by con- 
struction the triangle ACC is isosceles, the angle ACB is 
therefore equal to ACC, whence ACB, ACB are together 
equal to two right angles. The angles ACB ACB are 
therefore, the supplements of each other, since we understand 
by the supplement of an angle that which must be added to this 
angle in order to make two right angles. 

17. Prob. 4. One side and two of the angles being given, to 
find the remaining sides. 

Ex. 1. Given the side AB of a triangle (fig. 14), 40 feet, 
the angle ABC 52°, and the angle BAC 70°; to find the re- 
maining sides. 

Ex. 2. Given one side of a triangle 600 yards, the 
angle opposite 65°, one of th^ adjacent angles 40° 30' ; re- 
quired the remaining parts. 

18. It is now-evident, that we may always construct a 
triangle, when three of its parts are given, provided that one 



12 PLANE TRIGONOMETRY. 

of them be a side. If, however, the given parts be two of 
the sides and an angle opposite one of them, and the side 
opposite the given angle be less, than the other given side, 
to determine the triangle , we must know whether the angle 
sought be acute or obtuse. 

19. From the preceding examples it is easy to see, that 
the solution, obtained by geometrical construction, will not 
on account of the imperfection of instruments, be precisely 
the truth, but only an approximation to it. It is then desira- 
ble to ascertain methods of calculating triangles by numerical 
operations, since by means of these the calculation may be 
carried to any degree of exactness at pleasure. 

SECTION III. 

Of the solution of Triangles by Arithmetical calculation. 

20. In the solution of a triangle, it is obvious, that the 
magnitude of the angles will always come into considera- 
tion. Thus in the triangle ABC (fig. 1), if it be required to 
determine the third side AC, the angle at B and the sides 
AB,BC being given, since the value of AC evidently depends 
in part upon the magnitude of the angle at B, it is manifest, 
that the magnitude of this angle must be made to enter into 
the computation. But the angle at B is measured on the 
arc of a circle, while the sides are right lines ; of conse- 
quence the parts, which are necessarily the elements of the 
calculation, are different in kind and do not admit of compar- 
ison with each other. In endeavoring then, by means of 
numerical operations, to determine the different parts of a 
triangle, we are embarrassed at the outset by the difficulty 
of introducing into the calculation the magnitude of the 
angles. 

21. To show the manner, in w r hich this difficulty may be 
removed, we suppose that the triangles ABC, ate (fig. 15) 



PLANE TRIGONOMETRY. 13 

are similar, and that the sides ab, be, ac, of the latter are 
known ; if then in the triangle ABC the side BC be given, it 
is evident from the nature of similar triangles, that the 
remaining parts of this last may be determined from those of 
the other by means of a simple proportion. 

Let it then be supposed, that we have determined, by geo- 
metrical construction, the ten triangles in Plate 2, the first 
having the angle at B 10°, the second 15°, the third 20°, and 
so on ; and, for the sake of simplicity, let these triangles 
each be right angled at C. If it now be proposed to solve 
the triangle abc (fig. 16), right angled at c, the side be 
73 feet, and the angle b 40°, it is evident, on inspecting the 
series of computed triangles, that the triangle No. 7 is simi- 
lar to the triangle proposed ; since these triangles are both 
right angled, and one of the acute angles in the one is equal 
to an acute angle in the other. By means then of the com- 
puted triangle in the series, the remaining sides in the pro- 
posed triangle may be determined. 

Thus to find ac, we have the proportion, as the side BC 
to AC in the computed triangle, so is the side be to ac in the 
triangle proposed, that is, as 90 : 72 : : 73 : 58.4 the side ac 
required ; and to find ab we have BC : AB : : be : ab, or 90: 
115 : : 73 : 93.3 the side ab. In like manner we may deter- 
mine, any right angled triangle whatever, provided that one 
of its acute angles be equal to an acute angle in one of the 
triangles of the series. It will be easy to see moreover, 
that we may form a more extensive series of right angled 
triangles, by making in the first triangle an angle of one 
degree, in the second an angle of two degrees, in the third 
of three and so on to eighty nine degrees inclusive ; in 
which case, there will always be found, among the triangles 
of this series, one similar to any right angled triangle, which 
may be proposed ; provided that one of its acute angles be 
an entire number of degrees.. 



14 



PLANE TRIGONOMETRY. 



In like manner, it is evident, we may so extend the series 
of computed triangles, that there will always be found, 
among the number, a triangle similar to the one, which 
shall be proposed for calculation ; of consequence, as in the 
preceding example, the parts of this last may be determined 
from the former by means of a simple proportion. 

22. Thus far we have supposed the series of triangles ac- 
tually constructed, and preserved as in Plate 2, for the cal- 
culation of other similar triangles ; such an arrangement, 
however, would obviously be attended with numerous in- 
conveniences, especially as the series should become more 
extended. In order therefore to prevent these inconvenien- 
ces, we may write in a table the different parts in the several 
triangles of the series, expressing those, which correspond 
in each by the same general term, and arranging them in a 
uniform order. 

23. To form a table, for example, of the triangles, which 
we have calculated Plate 2, we may write down in order in 
the first column, as below, the angles at B in the several tri- 
angles, in the second the hypothenuses, in the third the 
bases, and in the fourth the perpendiculars, recollecting that 
by the term base we understand the side adjacent the angle 
at B in each of the triangles, and by the term perpendicular 
the side opposite this angle. 



Deg. 


1 Hyp. 


| Base 


| Perp. 


10 


242 


237 


43 


15 


127 


123 


33 


20 


121 


113 


42 


25 


139 


125 


59 


30 


132 


114 


65 


35 


125 


104 


70 


40 


115 


90 


72 


45 


103 


73 


72 


50 


93 


61 


69 


55 


132 


73 


109 



PLANE TRIGONOMETRY. 15 

24. To apply the table ; let it be supposed, that we know 
in the right angled triangle ABC (fig 17) the angle at B 35°, 
and the side BC 200 feet ; in order to solve this triangle, 
we look along the column of degrees for 35, against this we 
find 104 the side of the tabular triangle corresponding to 
the given side in the triangle proposed ; then, as 104 the 
base in the tabular triangle : 125 the hypothenuse of the 
same triangle : : BC 200 feet : 240.3 feet the hypothenuse 
of the triangle proposed. And as base 104 : perp. 70: : 200: 
134. 6 the side AC of the proposed triangle. 

SECTION IV. 

Of the Method of forming a Series of Triangles, having 
angles of every possible magnitude. 

Qo. Having seen the use, which may be made of a series 
of triangles containing angles of every possible magnitude, 
and the sides of which are calculated ; we next inquire into 
the method of constructing such a series. For the sake of 
simplicity let the triangles to be determined be right angled. 
It is evident (fig. 18), that such a series of triangles may be 
constructed in the quadrant of a circle. For if from each 
point in the arc AB, we let fall the perpendiculars ED, E'D' 
E"D", &c. upon the radius AC, and draw the radii EC, E'C, 
E"C, &c.,the triangles EDC E'D'C,E"D"C,&c. thus formed 
are right angled at D, D', D" &c, and the angles ECD, 
E'CD', E"CD" &c. will be successively of every possible 
magnitude. And since any two right angled triangles are 
similar, when an acute angle in one is equal to one of the 
rente angles in the other, there can not be proposed a right 
angled triangle, which will not be similar to some one of the 
triangles constructed in this manner. 

26. To form a table from the triangles of this series, it 
will be necessary, as before to apply definite terms to ex- 



16 PLANE TRIGONOMETRY, 

press their sides. The hypothenuses CE, CE', CE", &c. 
being all radii of the same circle, may be designated by the 
term radius. The perpendiculars ED, E'D', E'D ', &c. 
manifestly increase with the angles ECD, ECD', E"CD", 
&c. 3 to which they are opposite, or which is the same thing, 
with the arcs AE, AE', AE'' &c.,by which these angles are* 
measured. On account of this connexion we designate the 
perpendiculars ED, E'D', E "D", kc, by the term sine. 
We then define the sine of an arc, the 'perpendicular let fall 
from one extremity of this arc upon the radius, which passes 
through the other extremity. 

In order to fix upon a term to express the remaining side, 
we now remark, that two arcs are said to be complements the 
\ of the other, when the sum, or difference of these arcs is equal 
to ih \ part of the circumference of a circle. From the 

definition already given of a sine, the line EF, considered 
with reference to the arc EB, is the sine of that arc ; but 
En is the complement of AE ; EF therefore is the sine of 
the complement of AE, and may for the sake of brevity be 
called its cosine. But the side CD of the triangle CDE is 
equal to EF, since these lines are parallel to each other, 
and are included between parallels ; CD may therefore be 
considered the cosine of AE, and CD' of AE' &:c.; whence 
the remaining side in the triangles CDE, CDE', Sec. may 
be designated by the term cosine ; and we say, that the cosine 
of an arc is the sine of the complement of that arc, and is equal to 
that part of the radius comprehended between the centre and the foot 
of the sine. 

-21 . From what has been said, it will now be recollected, 
that in constructing, in the quadrant of a circle, a series of 
right angled triangles having angles of every possible mag- 
nitude, the radius of the quadrant forms the common hypoth- 
enuse in the several triangles, and that the remaining sides 



PLANE TRIGONOMETRY. 17 

are respectively the sines and cosines of the acute angle, 
which has its vertex at the centre. 

SECTION V. 

Of the Method of calculating the sides in the Series of Trian- 
gles, constructed in the Quadrant oj a Circle. 

28. Having ascertained a convenient method of construc- 
ting a series of triangles, in which the angles are of every 
possible magnitude, we proceed to investigate rules for the 
calculation of the sides. Since in the triangles CDE, CD'E', 
CD"E" it will be sufficient to know the ratio, which the 
sides have among themselves, it is evident, that we shall not 
have occasion to calculate the absolute, but merely the rela- 
tive value of these sides. We may therefore, for the sake of 
simplicity, consider the radius as unity and proceed to deter- 
mine the values of the sines and cosines in decimal parts of 
unity, or which is the same thing, we may consider the radi- 
us as divided into 100000 equal parts, and then determine 
the number of these parts in each of the sines and cosines. 

29. It is evident from inspection (fig. 18), that the sine of 
90° is equal to radius. And since (fig. 19) the side of an in- 
scribed hexagon is equal to radius (Geom. 271); if through 
the centre E of the arc BC we draw the radius EF, the chord 
BC will be bisected at G, and BG the sine of the arc BE 
will be equal to half of radius ; but the arc BE is equal to one 
third of a quadrant, or thirty degrees ; there are then two of 
the sines in a quadrant, the values of which we may know 
at the outset, viz. the sine of 90° equal to radius, and the 
sine of thirty degrees equal to one half of radius. 

30. Beginning with the sine BG (fig. 19), we find imme- 
diately its cosine GF j for in the triangle BGF, which con- 

3 



13 PLANE TRIGONOMETRY. 

tains these sides, and the hypothenuse of which is equal 
to radius or unity, we have BF 2 = BG 2 -f- GF 2 ; whence 
GF= V BF 2 — BG 2 , that is, cos EB= V R 2 — IhTEB 5 ! 

To find then the cosine of an arc, when the sine is given ; 
from the square of the radius subtract the square of the sine, the 
square root of the remainder will be the cosine. 

31. From what has been said, the relative value of the 
sides in one of the triangles of the series, which we wish to 
calculate, may now be considered as known ; we are next 
led to inquire into the use, which may be made of these in 
the calculation of the remaining triangles of the series. 

32. If the arc AE (fig. 20) be divided into two equal 
parts by the radius CD, the chord AGE will also be divided 
into two equal parts, and EG will be the sine of the arc DE 
one half of AE. The triangle AEF, right angled at F, gives 



AE _ ^AF 2 -^ EF 2 ; but A F= AC — FC = R — cos AE, 
and EF = sin AE ; squaring these values of AF and EF 
and substituting the squares in the expression for AE 

we have AE= V sin AE 2 +R 2 — 2RX cos AE + cos AE 2 ; 
butEF 2 4-FC 2 = CE 2 ,orsin AE 2 + cos AE 2 = R 2 , hence 
AE = s/ 2 R 2 — 2 RXcos AE ; whence \ the chord AE, or 



EG=|a/ 2R2 — 2RXcos AE. But EG is the sine of one 
half the arc AE ; whence to find the sine of half any arc less 
than a quadrant ; from twice the square of radius subtract twice 
the radius multiplied by the cosine of the given arc ; one half of 
the square root of the remainder will be the sine of half this arc. 

33. Let it now be supposed that we wish to calculate a 
series of triangles, in which there will always be found a 
triangle having for one of its acute angles any possible num- 
ber of minutes, or in other words, that we wish to calculate 
a series of triangles for every minute in the quadrant. By 



PLANE TRIGONOMETRY. 19 

art. 28 we have radius equal to 1.00000, or, omitting the sep- 
aratrix, to 100000 ; the sine of 30° will then be equal (art. 
29) to 50000, and its cosine (art. 30) to 86603. By the 
rule last obtained we may now calculate the sine of 15°, one 
half of 30°, the sine of 7° 30', one half of 15°, &c . Proceed- 
ing in this manner we obtain at the eleventh division, as 
below, the sine of 52" 44'" 3"" 45'"" equal to 00025, or 
preserving in the result one more decimal figure, to 000255. 

Sine Cos . 
30° 50000 86603 

15° 

7° 30' 
3° 45' 







06540 


99786 










00409 


99999 










00025 





0° 14' 3" 45'" 



0° 0' 52" 44'" 3"" 45'"" 
We next inquire into the manner, in which the sine of one 
minute may be obtained from this last. 

34. It is evident (fig. 21), that if )the arcs A6, Ac are 
very small, they will not differ materially from right lines ; 
the triangles A b d, A c e may then be considered similar, 
and A6 is to Ac, as bd to ce. In very small arcs therefore the 
ratio of the arcs is so nearly equal to the ratio of their sines, that 
the one may be taken for the other without sensible error. j" 

To find the sine of one minute from the sine of 52" 44'" 
3"" 45'"", we have then the following proportion, 52" 44'" 
3"" 45"'"; 60" : : 000255 : 00029 the sine of one minute. 

35. The sine and cosine of one minute may now be con- 

t The error in supposing that arcs less than one minute are proportional 
to their sines, will not affect the first ten places of decimals. 



£0 PLANE TRIGONOMETRY. 

sidered as known, and of consequeuce all the parts in the 
first triangle of the series, which we wish to calculate. We 
proceed to the investigation of principles, by which the re- 
maining triangles of the series may be computed from this 
first. 

Let it be supposed, that the sines of the two arcs AB. AC 
(fig. 22) are known ; it is proposed to find the sines of the 
sum and difference of these arcs. To construct the figure^ 
make the arc AD equal to the arc AC ; draw the chord CD 
and the radius LA ; the radius LA will divide this chord 
into two equal parts at the point I Geom. 106 ; from the 
points C. A, I and D. let fall upon BL the perpendiculars 
CK, AG. IH and DF : lastly, at the points I and D. draw 
IM and D>~ parallel to BL. Then CK ? the sine of BC, the 
sum of the two arcs AB ? AC is equal to EAI — 3IC ; but 
K3I is equal to HI ; CK is therefore equal to HI— 3IC, 
and the sum of the values of HI and 31 C will be equal to the 
sine of the sum of the two arcs AB. AC. To find the value 
of HI. we have LA : LI : AG ; HI. or putting for these lines 
what they severally denote R ; cos AC : : sin AB : HI, 

sin AB X cos AC 
hence HI = — 

And since the triangles LAG. CIM are similar, to find MC 
we have LA : LG : : CI : MC. that is, R : cos AB : : sin AC 

sin cos AB 

: MC. and MC = 

R 
Adding together these two expressions, we have 

sin AB ■ cos AC — sin AC X cos AB 

sin AB — AC = — 

R 

Then since CD is divided into two equal parts at the point 

I, C_\ is also divided into two equal parts at the point 31. and 

DF the sine of BD, the difference of the two arcs AB, AC 



PLANE TRIGONOMETRY. 21 

is equal to KN, that is to KM — MC, or which is the same 
thing to HI — MC. To find then the sine of the difference 
of the two arcs AB, AC we subtract the one from the other 
the values of the lines HI, MC already obtained, and 

sin AB V cos AC —sin AC X cos AB 

sin (AB — AC) = - - 

Having calculated therefore the cosines of any two arcs, 
the sines of which are given, in order to find the sine of the 
sum and of the difference of these arcs ; we multiply the 
sine of the first by the cosine of the second, and the sine of the sec- 
ond by the cosine of the first ; the sum of the two products divided 
by radius will be the sine of the sum of the two arcs ; and the dif- 
ference of the same products divided by radius will be the sine of 
the difference of these arcs. 

36 . The sine of the sum and of the difference of any two arcs 
being calculated by the rule now obtained, the cosines may 
be calculated by art. 30. When, however, the sines of any 
two arcs AB, AC (fig. 22) are given, the cosine of the sum. 
and of the difference of these arcs may be calculated in a 
more convenient manner. 

Since DC (fig. 22) is divided into two equal parts in I, 
FK will be divided into two equal parts in H, and LK, the 
cosine of BC, the sum of the arcs AB, AC will be equal to 
LH — HK, or to LH — IM. To find LH, we have LA : LI 
: : LG : LH, that is, R : cos AC : : cos AB : LH, and 

cos AB X cos AC 

LH = 

R 

To find IM, the similar triangles LAG, CIM give LA : 

AG ; : CI : IM, that is, R ; sin AB : : sin AC : IM, and 

sin AB X sin AC 

IM= - 

R 

3* 



22 PLANE TRIGONOMETRY. 

Subtracting this last expression from the former, we have 

cos AB X cos AC — sin AB X sin AC 

cos (AB + AC) = = = 

R 

Moreover LF, the cosine of BD, the difference of the arcs 
AB, AC is equal to LH-j-HF. or since HF is equal to 
IM, to LH -f- IM. To find therefore the cosine of the dif- 
ference of the arcs AB, AC, we add together the values of 

LH and IM already found, and 

cos AB X cos AC 4- sin AB > sin AC 
cos (AB — AC)= — ! 

To find therefore the cosine of the sum, and of the differ- 
ence of any two arcs, whose sines are given, having calcu- 
lated the cosines of these arcs, we take the product of the cosines, 
and also of the sines ; the difference of these two products divided, 
by radius will be the cosine of the sum of the two arcs, and the 
sum of the same products divided by radius will be the cosine of 
their difference. 

37. From the equation (art. 35) sin (AB-f-AC) = 

sin AB X cos AC -f sin AC X cos AB 

, we may find the 

R 
sine of twice any arc, whose sine and cosine are given ; 

for, as the arcs AB, AC may be of any dimensions, they 

may be considered equal ; substituting then AB for its 

equal AC throughout, we have 

2 sin AB X cos AB 

sin 2 AB = 

R 

To find then the sine of twice any arc, whose sine and cosine 

are given ; we multiply twice the sine of this arc by its cosine, and 

divide the product by radius. 



PLANE TRIGONOMETRY. 23 

38. From the equation (art. 36), cos (AB -f- AC) = 

cos AB X <^os AC — sin AB X sin AC 

1 . ., we may likewise 

R 

find the cosine of twice any arc, when its sine and cosine 
are known ; for supposing the two arcs equal, as before, 
and substituting AB for its equal BC throughout, we have 

cos AB 2 — sin AB 2 . 
cos 2 AB === j£ 

Hence, when the sine and cosine of an arc are known, to 
find the cosine of twice that arc ; we subtract the square of the 
sine of the arc from the square of its cosine, and divide the re- 
mainder by radius. 

39. We have seen (art. 34) the manner in which the sine 
of one minute may be calculated, consequently, all the parts 
in the first triangle of the series, which we wish to com- 
pute, may be considered as known. By the rules (art. 37, 
and 38) for the sine and cosine of twice any arc, we may 
now obtain from this first triangle the sine and cosine of 
two minutes, or the second triangle, and by the repeated 
application of the rules (art. 35, and 36) the remaining tri- 
angles of the series may be calculated. 

Ex. From the data given below it is required to calculate, 
the sine and cosine of 10°, 11°, &c, extending the calcula- 
tion to six places of figures, that is, supposing the radius to 
be divided into 1000000 equal parts. 

Sin. Cos. 

1° 017452 999848 
5° 087156 996195 



10° 

IP 

12° 

13° 224951 974370 

14° 

15° 



24 PLANE TRIGONOMETRY. 

40. We now perceive the advantage of constructing the 
series of triangles, which we wish to compute, in the quad- 
rant of a circle; since, if the radius of the quadrant be consid- 
ered as unity, the value of one of the sides in a triangle of 
the series will be known, and from this, radius being com- 
mon to all the triangles, the remaining sides in each of the 
triangles of the series may be determined. 

41. Having calculated a series of triangles by the rules, 
which we have now obtained, the different parts of this se- 
ries may be arranged in a table as in art. 23. The radius, 
however, which is common to all the triangles may be omit- 
ted, or written once only at the head of the table. 

SECTION VI. 

Of the solution of Triangles by the Table of Sines and 
Cosines. 

42. To apply the table of sines and cosines, let there be 
given for solution the right angled triangle ABC (fig. 23), 
in which the angle at B is 30°, and the hypothenuse AB 
three yards. 

Since we have considered the radius of the table as unity 
without reference to any particular denomination of meas- 
ure, it may manifestly, in all cases, be regarded as a unit of 
that species, in which the sides of a proposed triangle are 
expressed. With an extent then equal to one yard, which 
in the present instance will be the radius of the table, 
draw from B the arc DF ; from D let fall DE perpendicular 
to BC ; then will DE be the sine, and BE the cosine 
of the angle at B, and DBE will be the triangle of the ta- 
ble, similar to the proposed triangle ABC. To find the side 
AC, we then have BD : DE : : BA : AC, that is, R : sin B : : 
BA : AC, and to find BC, BD : BE : : BA : BC, that is, R : 
cos B : : BA : BC. Since the angle at A is the complement 



PLANE TRIGONOMETRY. 25 

of the angle at B, the cosine of B is the same with the sine 
of A ; these two proportions may therefore be united into 
one, and enunciated as follows ; radius is to the sine of one of 
the acute angles in a right angled triangle, as the hypothenuse to 
the side opposite this angle. 

43. Let it be proposed to solve the right angled triangle 
CDE (fig. 24). 

1. Given the hypothenuse CD 250 feet, and the angle at 
D 30° ; required the remaining parts. To find CE, we 
have 

R : sin D : : CD : CE 
or 100000: 50000: :250 ft.: 125 ft. 
And to find DE, 

R ; sin C : : CD : DE 
or 100000 : 86603 : : 250 ft.: 216.5075 ft. 

2. Given the hypothenuse CD 250 feet, and the side CE 
125 feet ; required the remaining parts. 

3. Given the side DE 216,5075 feet, and the angle C 
30° ; to find the remaining parts. 

44. From the preceding example, it is evident, that we 
may determine, by means of the table of sines and cosines, 
one of these three things, when two of them are given, viz. 
the hypothenuse, a side and one of the acute angles. 

SECTION VII. 

Of the second Method of constructing and calculating a Series 
of Triangles, having angles of every possible magnitude. 

45. Let there be given the two sides DE, EC in the tri- 
angle CDE (fig. 24), to find the hypothenuse, and one of 
the acute angles. It will soon be perceived, that the case 
now proposed cannot be solved by the table of sines and co- 
sines. This table, indeed, may contain a triangle similar to 



26 PLANE TRIGONOMETRY. 

the triangle proposed ; from the data now given, however, 
no part of that triangle can be known, except the radius or 
hypothenuse ; but neither of the sides DE, EC can be com- 
pared with radius ; of consequence from the parts now giv- 
en the remainder cannot be determined. 

46. For the solution of this case, it is then evident, that 
a series of triangles must be so constructed, that the radius 
shall be a side. In order to construct such a series of trian- 
gles, from the extremity of the radius AC (fig. 18) we raise 
the indefinite tangent AH, and draw from the centre C 
through each point in the quadrant AB the secants CG, CG', 
CG", &c. It is evident, that the triangles CAG, CAG', 
CAG", &c, right angled at A, must have all the combina- 
tions of angles, which can exist in a right angled triangle. 
The side AC moreover, common to all the triangles, will be 
the radius of the quadrant. 

47. In the application of terms to express the different 
sides in the triangles of such a series, the common side AC, 
for the same reason as in the former case, may be called ra- 
dius* The portions AG, AG', AG", &c, of the indefinite 
tangent AH, or in other words the perpendiculars in the tri- 
angles CAG, CAG', CAG", &c. are the tangents of the arcs 
AE, AE', AE", &c, since we define the tangent of an arc, 
that part which is intercepted on the tangent drawn through one 
extremity of this arc, by the two radii, which terminate it. The 
terra tangent may therefore be applied to designate these 
perpendiculars. The hypothenuses of the same triangles are 
secants of the arcs AE, AE', AE", &c, since we define the 
secant of an arc, the radius, drawn through one extremity cfthis 
arc and produced, until it meets the tangent, drawn through 
the other extremity. The hypothenuses of the triangles may 
in consequence be called secants. 

48. We proceed to develop rules for the calculation of 



PLANE TRIGONOMETRY. 27 

the sides in this second series of triangles. The triangles 
CDE, CAG (fig. 18), having the angle C common, are sim- 
ilar ; the sides of the latter may therefore be deduced from 
those of the former. 

To find AG, we have CD : DE : : CA : AG, that is, cos 
AE : sin AE : : R : tang AE, hence 

R X sin AE 
^8^= cosAE 

In order therefore to find the tangent of an arc, when 
the sine is given ; ice multiply the radius by the sine of the arc, 
and divide the product by its cosine. 

To find CG, we have CD : CE : : CA : CG,that is, cos 
AE : R : : R : sec AE, hence 

R 2 

sec AE = 



cos AE 

To find therefore the secant of an arc, the cosine being 
given ; we divide the square of the radius by the cosine of the arc. 

49. From what has been said, it is evident, that having 
calculated a series of triangles in the quadrant of a circle, in 
which the radius is the hypothenuse of the triangle, we may 
by the rules now obtained calculate a second series of trian- 
gles similar to the former, in which the radius shall be a side. 
The parts in this second series may, moreover be arranged in 
a table, as before, or which is the same thing, the table, al- 
ready formed from the first series of triangles, may be exten- 
ded by writing down against the several angles in this table 
the tangents and secants of these angles. 



28 PLANE TRIGONOMETRY. 

SECTION VIII. 

Of the solution of Triangles by the Table of Tangents and 
Secants. 

50. To show the use, which may he made of the table of 
tangents and secants, from the point F (fig. 23) draw the 
perpendicular FG ; then will FG he the tangent, and BG 
the secant of the angle B ; BF will he radius, and GBF the 
tabular triangle similar to ABC. 

To find AC, we have BF : FG : : BC : AC, that is, R : 
tang B : : BC : AC ; a proportion, which we may thus enun- 
ciate *, radius is to the tangent of one of the acute angles in a right 
angled triangle, as the side of the right angle adjacent to this 
acute angle is to the side opposite . 

51. Ex. 1. In the triangle ABC (fig. 25) given the side 
BC 500 yds, and the angle at B 22° 30'; required the side 
AC. 

To find AC, R : tang B : : BC : AC 

or 100000 : 41421 : : 500 : 207.105 
2. Given AC 207.105, and BC 500 yds, to find the angle 
at B. 

By means of the tangents then, we may determine one of 
these three things, when two of them are known, viz. the two 
sides of a right angle and an acute angle. 

52. It remains to show the use, which may he made of 
the secants. In the triangle ABC (fig. 23) we have BF : 
BG ; : BC : AB, that is, R : sec B: ; BC : AB, a proportion, 
which we thus enunciate ; radius is to the secant of one of the 
acute angles in a right angled triangle, as the side of the right 
angle adjacent to this acute angle is to the hypothenuse. 



PLANE TRIGONOMETRY. 29 

Ex. 1. In the triangle ABC (fig. 25) given BC 500 yds, 
and the angle B 22° 30', to find AB, 

R : sec B: ; BC : AB 
100000 : 1.08239 : : 500 ; 541.195. 

2. Given AB 541.195 yds, and the angle B 22 Q 30', to 
find BC. 

3. Given AB 541. 195, and BC 500 yds,to find the angle 
B. 

By means of the secants the same things are determined, 
as by the sines and cosines ; the secants are in consequence 
but seldom employed in the solution of triangles. 

53. The preceding examples, it will be perceived, do not 
comprehend the case, in which any two of the sides are given 
to find the third. This case, however, may be solved by 
means of the known property of aright angled triangle, viz. 
the square of the hypothenuse is equal to the sum of the 
squares of the two sides. It may moreover be resolved with 
facility by means of the two propositions art. 42, and 50. 

Ex. 1. In the triangle CDE (fig. 24) given the side CE 
274.703, the side D E 589.101, to find by means of the table 
the hypothenuse CD. Ans. 650. 

Ex. 2. In the triangle CDE ^fig. 24) given the hypothe- 
nuse CD 745, the side CE 427.317, to find, in like manner, 
the side DE. Ans. 610 

54. From what has been done, it is manifest, that any 
right angled triangle whatever may be determined by means 
of a table constructed in the manner, which has been des- 
cribed, provided that the triangle proposed be similar to 
some one of the triangles of the table, and that two of its 
parts be given, exclusive of the right angle. 



30 PLANE TRIGONOMETRY. 

SECTION IX. 

Of Cotangents and Cosecants. 

55. For the sake of convenience the table now constructed 
is usually still further extended, by writing down against 
the several angles in the table the tangents and secants of 
their complements. 

56. If from the extremity of the radius BC (fig 26), we 
draw the tangent BF to meet the secant CG, then, since the 
arc EB is the complement of the arc AE, BF will be the 
tangent, and CF the secant of the complement of the arc 
AE; BF may therefore for the sake of brevity, be called 
the cotangent, and CF the cosecant of AE, 

57 . The cotangent and cosecant of the arc AE (fig.26) evi- 
dently belong to a triangle distinct from that, in which the 
tangent and secant are found ; the former of these, however, 
may be deduced from the latter. Indeed, on account of the 
similar triangles CAG, CBF, we have GA : AC : : CB : BF, 
that is, tang AE : R: : R : cot AE, hence 

T> 2 

cot AE = — 

tang AE 

In order to find the cotangent of an arc, when its tangent 
is given;, we divide the square of the radius by the tangent of the arc. 

Again, to find FC, we have GA ; GC : : BC : FC, that is, 

tang AE : sec AE : : R : cosec AE, hence. 

sec AE X R 

cosec AE = 

tangAE 

To find therefore the cosecant of an arc, when its tangent 
and secant are given ; we multiply the secant of the arc by ra- 
dius, and divide the product by its tangent. 



PLANE TRIGONOMETRY. 31 

58. To show the use, which may be made of the cotan- 
gents and cosecants, let there be given in the triangle ABC 
(fig. 27) the side AC 6 yds, the angle at B 40°, to find the 
remaining side and the hypothenuse. From the point A with 
an extent equal to one yard describe the arc MK ; draw 
HM perpendicular to AC ; AHM will be the tabular triangle 
similar to the triangle proposed. 

1. To find BC we have 

AM : HM : : AC : BC 
R : Cot B : : AC : BC 
100000 : 1.19175 : 6 : 7.1505 
2. To find AB, AM ; AH : : AC : AB 

R : Cosec B ; : AC : AB 
100000 : 1.55723 : : 6 ; 9.3433 
3. Given AC 6, and AB 9.3433, to find the angle B. 

59. In employing the cotangent and cosecant of B in the 
preceding example, it is evident, that we have merely used 
the tangent and secant of A ; the work is of consequence the 
same , as if we had found by subtraction the angle A, and 
employed the tangent and secant of this angle. 

SECTION X. 

Miscellaneous remarks on the Trigonometrical Tables. 

60. In the preceding articles we have shown the manner, 
in which a set of trigonometrical tables may be formed. Va- 
rious methods of abridging the labor of calculation will natu- 
rally occur in practice. It is sufficient at present to remark, 
that from the nature of sines and cosines, tangents, and 
cotangents &c. if the sines, cosines, tangents &c. have been 
calculated to 45° inclusive, the remainder of the series may 
be considered as known. 

61 The several triangles of the tables for a given angle 



32 PLANE TRIGONOMETRY, 

may be exhibited in connection with each other. In the 
triangle ABC fig. 28), if we consider AC radius, BC will 
be the cotangent and AB the cosecant of the angle at B ; if 
then with an extent equal to AC we draw the arc DF, and 
from the points D and F the lines DE, FG perpendicular to 
BC, then FG will be the tangent and BG the secant of the 
angle at B ; DE moreover will be the sine and BE the co- 
sine of the same angle, and the triangles ABC, GBF, DBE 
will be the tabular triangles for the angle at B. 

62. The radius it will be observed, assumes a different 
position in each of these triangles ; in the triangle DBE, 
which contains the sine and cosine of the angle at B, it 
forms the hypolhenuse ; in GBF, which contains the tangent 
and secant of the angle at B, it becomes the side adjacent to 
this angle, and in ABC which contains the cotangent and 
cosecant of the same angle it becomes the side opposite. 
This circumstance then must evidently be taken into consid- 
eration in calculating a proposed triangle by means of the 
tables. If the hypothenuse of the proposed triangle be 
made radius, or if it be compared with the radius of the 
tables, the sines and cosines must be employed in the calcula- 
tion of this triangle ; if the side adjacent a given or required 
angle be made radius, the tangents and secants must be em- 
ployed in the calculation ; and if the side opposite the given 
or required angle be made radius, the cotangents and cosecants 
must be employed. 

63. The numbers in the tables, the construction of which 
has now been explained, are usually denominated natural 
sines, tangents, &c. By means of these numbers, it has been 
seen, that the sides and angles of any proposed triangle, 
which is similar to some one of the triangles of the tables, 
may be accurately determined ; the calculations, however, 
must be made by the tedious processes of multiplication and 



PLANE TRIGONOMETRY. 33 

division. To avoid this inconvenience, another set of tables 
may be constructed by writing down against each of the 
angles the logarithm of its natural sine, tangent, &c. ; of 
consequence, in the use of the numbers of such a set of tables, 
addition and subtraction will take the place of multiplication 
and division. 

64. The logarithmic sines, tangents, &c. are commonly 
called artificial sines, tangents &c. to distinguish them from 
the natural sines, tangents, &c. In comparing these two sets 
of tables, one circumstance will require consideration. The 
radius, to which the natural sines, &c. are calculated, is uni- 
ty. The secants and a part of the tangents are therefore 
greater than a unit ; while the sines and another part of the 
tangents are less than a unit ; of consequence, when the log- 
arithms of these are taken, some of the indices will be posi- 
tive, and others negative. To remedy therefore the inconven- 
ience, which would result from the occurrence of both posi- 
tive and negative indices in the same set of tables, ten is 
added to each of the indices, by which they are all rendered 
positive. Thus the natural sine of one minute is 0.00029, or 
more accurately 0.000290888; the logarithm of this is 4^46373; 
but the index by the addition of 10 becomes ( 10 — 4) or 6 ; 
of consequence the logarithmic sine of one minute is 6.46373. 
In like manner the logarithmic sine of two minutes is 
6 .76476, and so of the rest. 

SECTION XL 

Of the solution of oblique angled Triangles. 

65. From what has been said, it is evident, that, in order 
to calculate a triangle by means of the tables, we must be 
able to construct from the numbers in the tables a triangle 
similar to the one, which is proposed for calculation. 



34 PLANE TRIGONOMETRY. 

Let there be given the oblique angled triangle ABC (fig. 
29, A), it is proposed to construct from the numbers in the ta- 
bles a triangle, which shall be similar to this. Having circum- 
scribed a circle about the given triangle, from the centre D 
of this circle draw the radii DA, DB, DC ; with the radius 
D6, equal to that of the tables, describe the circle abc ; draw 
the chords ab, 6c, ac joining the points of section a, b and c ; 
since Da is equal to D6, the lines DA, DB are cut propor- 
tionally at the points a and b ; ab is therefore parallel to AB 
(Geom. 199) ; for a similar reason, ac is parallel to AC, and 
be to BC ; the triangle abc is therefore similar to the trian- 
gle ABC (Geom. 209). Upon the side ab of the triangle abc 
let fall from D the perpendicular De ; since aD is equal to 
the radius of the tables, ae will be the sine of the angle «De, 
and be the sine of the angle 6De ; produce De to meet the 
circumference of the circle abc in h ; since the arc ah is equal 
to the arc hb (Geom. 105), the angle aDe is equal to the an- 
gle 6De ; the side ab of the triangle abc is therefore equal to 
twice the sine of the angle aDe, or since the angle aDe is 
equal to the angle ac& (Geom. 126), the side ab is equal to 
twice the sine of the angle acb ; in like manner be is equal 
to twice the sine of the angle at a, and ac to twice the sine 
of the angle at b ; but the angles a, b and c in the triangle 
a6c are equal respectively to the angles A, B and C in the 
triangle ABC ; the sides a&, 6c, ac in the triangle abc are 
therefore equal respectively to twice the sine of the angles 
C, A and B opposite the homologous sides AB, BC, AC in 
the triangle ABC. 

With respect to the side ab in the case, where the 
angle C is greater than a right angle (fig. 29, B) ; since 
from the definition of a sine, an angle will have the same 
sine with its supplement to two right angles, ae, the sine 
of the angle aDe, will also be the. sine of the angle aDk : 



PLANE TRIGONOMETRY. 35 

and the side ab will be equal to twice the sine of the angle 
aDh; but the angle aDk, or, which is the same thing, the an- 
gle ADE is equal to the angle ACB; the side ab is therefore 
equal, as before, to twice the sine of the angle at C. From 
the numbers in the tables we may then always construct a 
triangle similar to any oblique angled triangle, which may 
be proposed for calculation, by taking for each of the sides 
in the required triangle twice the sine of the angle opposite 
the homologous side in the triangle proposed. The tables, 
therefore, which at first appear limited to the solution of 
right angled triangles, may evidently be applied to the solu- 
tion of triangles of whatever kind. 

66. We proceed to the investigation of rules for the calcu- 
lation of plane triangles of whatever kind by means of the 
tables. The similar triangles ABC, abc (fig. 29) give AB : 
BG : : ab : be &c, that is, AB : BC : : 2 sin C : 2 sin A, or, 
sin C : sin A ; a proportion, which, it is easy to see, is of 
general application, and which may be enunciated, as fol- 
lows ; in any triangle whatever, the sines of the angles are to 
each other as the sides opposite to these angles. 

67. Case 1 . To apply this principle, let there be given 
in the triangle ABC (fig. 30) the side BC 70 yds, the angle 
A 86°, and the angle C 45°, to find the remaining parts* 

1. To find AB, sin A 86° 9.99894 

BC 70 yds 1.84510 

sin C 45° 9.84949 

TlT69459 



AB 49.62 yds U69565 

2. In like manner, AC is found equal to 52.96 yds. 

Case 2. Given two of the sides, viz. AB 49.62 yds, BC 
70 yds, and the angle A opposite BC 86° ; to find the re- 
maining parts. 



35 PLANE TRIGONOMETRY. 

1. To find the angle at C. 

BC70yds 1.84510 

sin A 86° 9.99894 

AB 49.62 yds 1.69565 

11769459 



sin C 45° 9.84949 

2. Having obtained the angle C, the side AC is found 
as in the preceding case. 

3. Given in the triangle ABC (fig. 13) the side AB 72 
feet, the side AC 41 .5 feet, and the angle B 30°, to find the 
angle C. 

AC 41.5 1.61805 

sin B 30° 9.69897 

AB 72 1.85733 

lT.55630 



sinC 60° 10' 9.93825 
In this last example the solution (art. 16) is ambiguous ; 
since, however, the angles ACB, ACB are supplements, the 
one of the other, and the sine is the same for an angle, and 
its supplement ; if the angle 60° 10' first obtained does not 
answer the conditions of the question, we then take its sup- 
plement 1 19° 50' for the angle C, which is sought. 

68. From the preceding examples, it appears, that we 
may always solve a triangle by the rule just obtained, 1. 
when a side and two of the angles are given ; 2 . when two of the 
sides and an angle opposite to one of them are given. 

69. In the triangle ABC (fig. 30) let there now be given 
the sides AC and BC, and the angle C contained between 
them, to find the remaining parts. This case, it will be 
perceived, does not admit of solution by the preceding rule; 
a new one must therefore be sought. Having described a 
circle with a radius equal to that of the tables (fig. 31), draw 
the diameter AM ; let the arc AB, set oft' from A, be equal 
to the number of degrees, contained in the angle at A in the 



PLANE TRIGONOMETRY. 37 

proposed triangle ; and also the arc AC to the number of de- 
grees in the angle at B ; from B draw the chord BD per- 
pendicular to AM ; from C draw CP perpendicular and CF 
parallel to AM ; join BF, FD ; from F, with the radius of 
the circle ABD equal to that of the tables, draw the arc IGK 
meeting CF in G, and, at the point G, draw HL perpendic- 
ular to CF ; the line GL is the tangent of the angle GFL, 
or which is the same thing, of CFD, and GH is the tangent 
of GFH, or CFB ; and the angles CFD, CFB having their 
summets in the circumference are measured by half the arcs 
CD, CB, on which they stand ; but CD is the sum of the 
two arcs AB, AC, and CB is the difference of these two 
arcs ; whence GL is the tangent of the half sum, and GH 
the tangent of the half difference of the arcs AB, AC ; but 
the arcs AB, AC, by hypothesis, are respectively the meas- 
ures of the angles A and Bin the proposed triangle ABC 
(fig, 30) ; GL is therefore the tangent of the half sum and 
GH the tangent of the half difference of the angles A and B ; 
in like manner it may be shown, that DE is the sum, and 
BE the difference of the sines of the same angles A and B ; 
wherefore on account of the parallel lines BD, HL we have 
DE : BE : : LG : GH, or putting for these lines what they 
severally denote 

A + B A — B 

sin A -j- sin B : sin A — sin B : : tang — ^ — : tang 5- — ; 

but by art. 66 we have sin A : BC : : sin B : AC, hence 

sin A + sinB:sin A— sin B : :BC+AC:BC — AC : 

A + B A — B 
whence BC -f- AC : BC — AC : : tang — g— : tang — -^ . 

a proportion, which we may thus enunciate ; the sum of two 
sides of a triangle is to their difference, as the tangent of half 
the sum of the opposite angles is to the tangent of half their differ*- 
<ence. 



33 PLANE TRIGONOMETRY. 

70. Ex.1. Let AC (fig.30) be 52.96 yds, BG 70 yds, and the 
angle C 45° ; it is required to find the remaining parts. 
Subtracting the angle C 45° from 180°, and dividing the re- 
mainder by 2, we have 

— i— = 67° 30', and to find AzJ^ we have 

2 2 

BC-j-AC 122.96 2.08976 

BC — AC 17.134 1.23147 

A+B 

tang — £ 67° 3 0' 10.38278 

TT61425 



A — B 

tang — £ 18° 30' 9.52449 

Having obtained the half difference between A and B, 
the greater angle A 86<* is found by adding the half differ- 
ence to the half sum, and the less angle B 49° by subtracting 
the half difference from the half sum. The angles being thus 
obtained, the remaining- side may be calculated as before. 

Ex. 2. Given two sides of a plane triangle 450 and 540, 
and the included angle 80°, to find the remaining parts. 

Ans. Angles 56° ll',43 Q 49', and side 640.08 

Ex. 3. Given two sides of a plane triangle 76 and 109, 
and the included angle 101° 30', to find the remaining parts. 
Ans. Angles 30° 57' 30", 47° 32' 30", and side 144.8. 

71. The preceding rules do not include the case, in 
which the three sides are given to find the angles. To ob- 
tain an expression for this case, in the triangle ABC (fig. 32) 
let fall from the point B, BD perpendicular to AC, and from 
the same point, with a radius equal to the side BC, describe 
the circumference CEHF ; extend the side AB, until it 
meets the circumference in E ; then, since AE and AC are 
drawn from the same point A without the circle, AC : AE :: 
AG : AF(Geom.225); but AE is equal to AB + BC, and AG 
is equal to AB — BC ; AF moreover is equal to AD — DC; 



PLANE TRIGONOMETRY. 39 

therefore AC : AB -f BC : : AB — BC : AD — DC . If then 
upon the longest side of a triangle we let fall a perpendicu- 
lar from, the opposite angle, we have the following propor- 
tion ; as the longest side of a triangle is to the sum of the two 
other sides, so is the difference of these last to the difference of the 
segments made by the perpendicular . 

72. Ex. 1. Given the three sides of a triangle ABC (fig.33), 
viz. AB 66 feet, AC 75, and BC 34 feet, to find the angles. 
By the preceding rule 

AC 75 1.87506 

AB+BC100 2.00000 
AB — BC 32 1.50515 



3.50515 



AD — DC diff. seg. 42.67 1.63009 

Adding half the difference of the segments to half the sum, 
we obtain the greater segment AD 58.83 feet ; subtracting 
the half difference from the half sum, we obtain the less 
segment DC 16.17 feet. 

Thsn to find the angle A, AB : R : : AD : cos A 26° 57' : 
and to find the angle C, BC : R : : DC : cos C 61° 36'. 

Ex. 2. The sides of a plane triangle are 40, 34 and 25 ; 
feet respectively ; required the angles. 

Ans. 38° 25' 20", 57° 41' 25", 83 Q 53' 15". 
Ex. 3. The sides of a plane triangle are 390, 350 and 270 
feet respectively ; required the angles. 

Ans. 42° 21' 57", 60° 52' 42", and 76° 45' 21" 



40 PLANE TRIGONOMETRY. 

SECTION XII. 

Mensuration of Heights and Distances. 

72. The preceding principles may be readily applied to 
the mensuration of the heights and distances of objects, 
in cases where a direct measurement cannot be made or 
where it cannot be made with convenience. The follow- 
ing are among the problems which most frequently occur in 

practice. 

PROBLEM I. 

73. To determine the altitude BC of a steeple or other 
object (fig. 37), situated on a horizontal plane. 

From the bottom of the steeple measure, in a direct line 
upon the horizontal plane, any convenient distance BA ; at 
the extremity A of this line measure the angle BAC, com- 
prised between the line BA and an imaginary line, drawn 
from the point A to C the top of the steeple ; then in the 
triangle ABC the side AB and all the angles will be known; 
whence to find BC, we have 

sin ACB : AB : : sin BAC : BC 
or R : tang BAG : : AB : BC 

The angle BAC situated in a plane perpendicular to the 
horizon, and comprised between a horizontal line AB and an 
ascending line AC, is called an angle of elevation. 

The line AB, which has been measured in order to deter- 
mine by means of it the required height, is called a base line. 

Ex. 1. Let the base line AB be 200 feet, and the angle 
of elevation BAC be 47 Q 30' ; required the height of the 
steeple. Ans. 218.26 feet. 



PLANE TRIGONOMETRY. 41 

Ex. 2. Let AB be 384 feet, and the angle at A 36° 52' ; re- 
quired the height of the steeple. Ans* 288 feet. 

PROBLEM II. 

74. To determine the perpendicular height of a cloud or 
other object above a horizontal plane. 

At two convenient stations A and B on the same side 
of the object (fig. 38) or on opposite sides (fig. 39) and in 
the same vertical plane, let two observers take at the same 
time the angles of elevation CAB, CBD, and let the dis- 
tance BAbe measured, then the exterior angle of a triangle 
being equal to the two interior and opposite, subtracting 
(fig. 38), the angle CAB from CBD, we have the angle ACB ;. 

whence sin ACB : AB : : sin CAB : BC 

then in the triangle CBD, we have to find CD the height 
required R : BC : : sin CBD : CD 

or (fig. 39) subtracting the sum of the angles at A and B 
from two right angles, we have the angle ACB; to find one 
of the remaining sides in the triangle ACB, AC for example,, 
we have sin ACB : AB : : sin ABC : AC 

then in the triangle ADC, to find CD we have 
R : AC : : sin CAD ; CD 
Ex. 1. Let the angles of elevation (fig. 38) be 31° and 
46° respectively, and the line AB 100 yards ; what is the 
height of the cloud ? 

We have in this example ACB = 15°, then 

sin ACB 15° 9.41300 

AB 100 2.00000 

sin CAB 31° 9.71184 

BC 2.29884 

Then Radius 10.00000 

BC 2.29884 

sin CBD 46^ 9.85693 

CD 143. 14 1U55TI 
5 



42 PLANE TRIGONOMETRY. 

Ex. 2. Let the angles of elevation (fig. 39^) be 53° and 
79° 12' respectively, and the distance AB 100 rods; what is 
the altitude of the cloud ? Ans. 105.89 rods. 

PROBLEM III. 

75. To find the height of an inaccessible object standing 
on a horizontal plane. 

At two stations A and B in the same vertical plane pas- 
sing through the top of the object (fig. 40), take the angles 
of elevation CAB, CBD, and measure the distance AB ; then 
ia the triangle CAB we have 

sin ACB : AB : : sin CAB : CB 

whence to find CD the altitude required, we have in the tri- 
angle CBD R : CB : : sin CBD : CD 

Ex. 1. Let the angles of elevation be 32° and 58°, and 
the base line 100 yards ; required the height of the object. 

Ans. 102.51 yards. 
Ex, 2. Let the angles of elevation be 40° and 60 Q and 
the base line 100 feet • required the height of the object. 

Ans. 162.75 feet. 
PROBLEM IV. 

76. To find the height of an accessible object standing 
on an inclined plane. 

From the bottom of the object (fig. 41) measure in the 
same direct line any two distances AB, BD, at B take the 
angle CBA, and at D the angle CDA, then in the triangle 
CDB we have sin DCB : DB : : sin CDB : BC 

BC being thus determined, we have in the triangle CBA 

BC + BA : BC — BA : : tang \ (A + C) : tang \ (A — C) 
having found by means of this proportion one of the re- 
maining angles in the triangle CBA, BCA for example, we 
have, to find the height required 

sin BCA: BA : : sin ABC : AC 



PLANE TRIGONOMETRY. 43 

Ex. Let the distances AB, BD be 40 and 60 feet, the 
angle at B 41°, that at D 23° 45' ; required the height of 
the object. Ans. 57.624 feet. 

PROBLEM V. 

77. To find the distance between two objects inaccessi- 
ble by reason of an intervening river, the objects being both 
on the same level. 

At two stations A and B (fig. 42) also on a level, we 
take the angles CAD, DAB, CBA, DEC and measure the 
distance AB ; then in the triangle ACB we have 

sin ACB : AB : : sin CAB : CB 

and in the triangle ADB 

sin ADB : AB : : sin DAB : DB 

then in the triangle CBD we have 

BC + BD: BC — BD: : tang £ (D + C):tang ±(D — C) 
having found by means of this proportion the angle BCD for 
example, we have finally for the distance sought 

sin BCD : BD : : sin CBD : CD 

Ex. Let the angles in the order stated above be 37°, 58° 
20', 53° 30', 45° 15', and the base line 300 yards. What is 
the distance between the two objects ? Ans. 479.79 yards. 

PROBLE3I VI. 

78. The height of a tower being given to find the hori- 
zontal distance between two objects, situated on the same 
level with the tower and in a direct line from the bottom of it. 

Take the angles EAD, EAC (fig. 43) contained between 
the line EA parallel to the horizon and lines drawn from A, 
the top of the tower, to the objects D and C respectively, then 
in the triangle DAB all the angles and the side AB will be 



44 PLANE TRIGONOMETRY. 

known, from which DB may be found ; in like manner in 
the triangle ACB all the angles and the side AB will be 
known from which CB may be found ; the value of these be- 
ing determined, the latter subtracted from the former will 
give the distance sought. 

An angle EAD comprised between a line parallel to the 
horizon and a descending line is called an angle of depression, 

Ex. Let the height of the tower be 120 feet, the angle of 
depression of the nearest object 57°, that of the most remote 
25° 30'. What is the distance between the two objects ; 

Ans. 173.65 {eet. 

PROBLEM VII. 

79. To find the altitude of a hill above a horizontal plane 
passing through a given point, on the side of another hill 
opposite. 

At the given point B (fig. 45) take the angle of elevation 
CBD of the top of the hill, the altitude of which is sought, 
from B measure a base line BA directly up the other hill, 
and at A take the angle of elevation CAE and the angle of 
depression EAB, then in the triangle CAB all the angles 
will be given and the side AB to find CB ; and CB being 
found, we shall have in the triangle CBD all the angles and 
the side CB to find CD the altitude sought. 

Ex. Let the angle CBD be 5° 52', the base line AB 642 
yards, and the angles CAE, EAB 3° 59', and 39' respective- 
ly. What is the altitude of the hill ? Ans. 161 .3 yards. 

PROBLEM VIIL 

BO. To find the distance between two inaccessible objects, 
both of which can be seen at the same time from one point 
only. 

Let D (fig. 46) be the point from which both objects can 
be seen at the same time; in any convenient directions take 



PLANE TRIGONOMETRY. 45 

a station, C where A can be seen, and a station E, where B 
can be seen, and measure the distances CD, DE ; at 
D take the angles ADC, ADB, BDE, and at C and E the 
angles ACD, BED ; then in the triangles ACD, BDE we 
shall be able to determine the sides AD, DB,and these being 
found, we shall have in the triangle ADB data sufficient to 
determine AB the distance sought. 

Ex. Let CD be 200 yards, the angle ADC 89°, ACD 
50° 30'; letDE also be 200 yards, the angle BDE 54° 30', 
BED 88° 30', and ADB 72° 30' ; required the distance be- 
tween the two objects. Ans. 345.5 yards. 

PROBLEM IX. 

81 . To find the altitude of an object, .situated on the sum- 
mit of a hill, above a horizontal plane at the foot of the hill. 

Having assumed a station B (fig. 47) upon the horizon- 
tal plane, measure in a direct line from the object a base 
line BA, at B take the angles of elevation DBE, CBE, and 
at A the angle of elevation CAE ; in the triangle CAB we 
determine the side CB, then in the triangle CBE we find 
the remaining sides CE and BE ; this being done, in the tri- 
angle DBE we find the side DE ; subtracting next DE from 
CE already found we have the altitude CD sought. CD may 
also, it is evident, be found by means of the triangle CBD 
without calculating CE and DE., 

Ex. Let the angle DBE be 40°, CBE, 51°, the base line 
BA 100 yards, and the angle CAE, 33° 45', to find the alti- 
tude of the object. Ans, 46.67 yards. 

PROBLEM X, 

82. To determine the distance of an inaccessible object, 
when an instrument for measuring angles cannot be procured. 

Let E be the object (fig. 50) and AE the distance sought, 
from the station A in the direction EA, measure any conven- 
ient distance AC ; assume another station B, and in the direc- 

5* 



46 PLANE TRIGONOMETRY. 

tion EB measure any convenient distance BD ; measure also 
the distances AB, AD and BC . In the triangles ACB, ABD 
we shall have the sides given to determine the angles, and 
these being found, those of the triangle AEB will be known ; 
then in the triangle AEB we have the angles and a side 1 to 
find AE the distance sought. 

Ex. Let AB be 500, AC 100, CB 560, BD 100, and 
AD 550 yards ; required the distance AE. 

Ans. 536.25 yards. 

QUESTIONS AND PROBLEMS FOR PRACTICE. 

1 . A line 27 yards long will exactly reach from the top of a 
fort to the opposite bank of a river, known to be 23 yards 
broad ; what is the height of the wall ? Ans. 42.42 feet 

2. The height of an object on a horizontal plane is 200 
feet, and its angle of elevation at the place of the observer is 
42° 30' ; what is his distance from the object ? 

Ans. 218.26 feet. 

3. A ladder 193.55 feet long is placed against a wall in 
an oblique position, and reaches to a point 75.83 feet from 
the ground ; what is the angle at which it is inclined to the 
ground ? Ans. 23° 3' 55". 

4. From the edge of a ditch 18 feet wide, surrounding a 
fort, I took the angle of elevation of the top of the wall and 
found it 62° 40' ; required the height of the wall, and the 
length of a ladder necessary to reach from my station to the 
top of it. Ans. Height 34.82, length 39.2 feet. 

5. A ladder 125 feet long is placed against a wall so that 
the angle at the bottom is double the angle at the top ; how 
high up the wall does it reach and how far distant from the 
wall is its foot ? Ans. 108.25, and 62.5 feet respectively. 

6. A person attempts to swim across a river, whose 
breadth is 329 yards. How far from the place immediately 



PLANE TRIGONOMETRY. 47 

opposite to his departure does he arrive, if the stream forces 
him to swim twice as far as he would have done, had there 
been no current ? Ans. 569.9 yards. 

7. From the top of a tower, 143 feet high, by the sea 
side, I observed that the angle of depression of a ship's bot- 
tom, then at anchor, was 55° ; what was its distance from 
the bottom of the wall ? Ans. 100.13 feet. 

8. Wanting to know the breadth of a river, I measured 
100 yards in a right line close by one side of it, and at each 
end of this line I found the angles subtended by the other 
end and a tree close by the other side of the river to be 53° 
and 79° 12'. What is the perpendicular breadth ? 

Ans. 105.89 yards. 

9. Being on one side of a river and wanting to know 
the distance to a house, which stood on the other side, I 
measured 200 yards in a right line by the side of the river, 
and found that the two angles at each end of this line form- 
ed by the other end and the house were 73 Q 15' and 68° 2'; 
what was the distance between each station and the house ? 

Ans. 296.54 and 306.19 feet. 

10. A gentleman wishing to ascertain the distance be- 
tween two trees A and JB, which could not be directly meas- 
ured on account of a pool, that occupied the intermediate 
space, assumed a station C, from which both could be seen, 
and found by measurement the distance AC 7.35 chains, the 
distance BC 8.4 chains, and the angle ACB 55° 40'. What 
is the distance of the trees A and B. Ans. 7.412 chains. 

11. From the top of a tower, whose height is 108 feet, 
the angles of depression of the top and bottom of a vertical 
column standing in the horizontal plane are found to be 30 Q 
and 60° respectively; required the height of the column. 

Ans. 72 feet. 

12. From a window B near the bottom of a house sup- 



48 PLANE TRIGONOMETRY. 

posed to be on a level with a monument CD, the angle of 
elevation CBD of the top of the monument being 40°, and 
from another window A 18 feet higher the angle of eleva- 
tion CAE being 37° 30', it is required to find the height of 
the monument CD and its distance BD. 

Ans. CD 210.44 feet, and BD 250.79 feet. 

13. Wanting to know the height and distance of an ob- 
ject on the other side of a river, and on a level with the 
place where I stood, close by the side of the river ; not hav- 
ing room to go backward on the same plane, on account of 
the immediate rise of the bank, I placed a mark where I 
stood and measured in a direct line from the object up the 
hill a distance of 132 yards ; I then found the angle of de- 
pression of the mark by the river's side 42°, that of the bot- 
tom of the object 27 Q , and of its top 19° ; required the height 
of the object, and the distance of the mark from its bottom. 

Ans. Height 28.63, distance 75.25 yards. 

14. It is required to find the height of a castle AB, sit- 
uated on an eminence by the sea shore, above the level 
of the sea and its horizontal distance from a ship S at an- 
chor, the angles of depression, EAS, FBS being given 
equal respectively to 4° 52', and 4 Q 2', and the height of the 
castle being 54 feet. 

Ans. Distance 3690 feet, and height 314 feet. 

15. At a station A at the bottom of a hill, I took the an- 
gle of elevation of the top of an object D on the summit of 
the hill, 38°, and measuring directly up the hill a distance of 
676.47 feet to another station B, 1 took at this station the 
angle of elevation of the top of the object D, 46°, and the 
angle of depression of the first station A, equal to the angle 
of elevation of the second station from the first, 27° 30' 
25" ; what is the height of the object D above the level of 
the first station A. Ans. 949.19 feet. 



PLANE TRIGONOMETRY. 49 

16. Suppose the object CD (fig. 45) to stand upon a hori- 
zontal plane ABC and that AB is equal to 250 yards, and 
that the angles at its extremities are known, viz. CAB 56° 
46', CBA 62° 54', DAC 6° 40', DBC 1° 6', What is the 
height CD and the two distances AD and BD ? 

Ans. AD 254.989, BD 238.814, and CD 29.745 yards. 

17. To ascertain the distance between two places, the 
angle which they subtended at a point equally distant from 
both was observed, and again at another point equally dis- 
tant from both ; and the distance between the points of ob- 
servation was measured. To determine from these data 
the distance of the places, 1° when the points of observation 
are both on the same side of the places, 2°, when they are 
on opposite sides. 

18. At a given distance from an obelisk, whose height is 
known, a colossal statue on the top of it subtends the same 
angle as tbe observer, when seen from the base of the obelisk. 
Supposing the obelisk and the observer on the same hori- 
zontal plane, and the height of the observer known, show 
how the height of the statue may be determined. 

19. Two stations on the side of a mountain are within 
sight of each other, of a signal D placed upon the summit of 
the mountain and of another E upon the horizontal plane ; 
supposing the distance between the two stations given, ex- 
plain the method of finding the height of the mountain by 
means of observations made at these two stations, 1° when 
the stations are in the same vertical plane with the two sig- 
nals^ when they are in different planes. 

20. Two spectators at two different stations on a hori- 
zontal plane, the distance between which is known, observe 
at the same instant a balloon, which rises into the air at a 
uniform rate ; and after a given interval of time again observe 



50 PLANE TRIGONOMETRY 

it ; what are the observations necessary to determine the 
two elevations of the balloon and its rate of moving, and 
how shall the data be applied to this purpose, 1°, when it as- 
cends in a perpendicular line and the stations are not in the 
same vertical plane with it, 2°, when it ascends in an oblique 
line and the stations are in the same vertical with it. 

SECTION XIII. 

Of the Line of natural Sines, Tangents, fyc. 

83. Let the quadrant AB (fig. 35) be divided into equal 
portions, of a degree each, for example ; and let lines be 
drawn, parallel to AC, from each point of division in the 
quadrant AB to the line CB. The several distances from C 
to the points, where these parallels meet the line CB, will 
be equal respectively to the lengths of the sines for each de- 
gree in the quadrant AB. Thus the extent from C to 10 will 
be equal to the sine of 10°, from C to twenty to the sine of 
20°, and so on. The line CB, divided in this manner, is 
called a line of sines. 

In like manner if through each of the points of division in 
the quadrant AB secants be drawn to meet the indefinite 
tangent BD, the several distances from B to the points of 
division in BD will be equal respectively to the tangents for 
each degree in the quadrant AB, and the line BD will be a 
line of tangents. 

If moreover the different distances from the centre C to 
each of the divisions on the line of tangents BD be transfer- 
red to the line CE, the line CE, will be a line of secants. 
Thus B 20 is the tangent, and C 20 the secant of 20°. 

84. A single example will show the use, which may be 
made of these lines, 



PLANE TRIGONOMETRY. 51 

In the triangle ABC (fig. 30), given the side BC 70 yds, 
the angle A 86°, and the angle C 45°, to find AB. 

By art. 66, sin A 86<* : sin C 45° : : BC : AB. In order 
therefore to determine the value of ABby the line of sines, 
we measure by means of a convenient scale of equal parts, 
the extent from C to 86 on the line of sines (fig. 35), and 
also the extent from C to 45 ; we then have the relative val- 
ues of the first and secoud terms in the above proportion. 

Let the sine of A determined in this manner be 332, and 
the sine of C 235, then 

332 : 235 : : 70 : 49.5 the side AB. 

The method of solving triangles, which has now been ex- 
hibited, is obviously the same, as that by the table of natural 
sines, &c, with the exception that the values of the sines in 
the present case are determined by geometrical construc- 
tion. 

SECTION XIV. 

Of the Line of Logarithmic Sines, Tangents, fyc. 

85. The logarithms of numbers may be conveniently 
represented by lines. Thus, since the logarithm of 10 is 1 
and the logarithm of 100 is 2, if a line of one foot ,be made 
to represent the logarithm of 10, a line of two feet will rep- 
resent the logarithm of 100. 

For the purpose of forming a logarithmic scale let the line 
ab (fig. 36), intended for the scale and taken of any length 
at pleasure, represent the logarithm of 100; then 100 will 
stand at the end of the scale ; and 1, the logarithm of which 
is 0, will stand at the beginning of the scale. For the in- 
termediate numbers 1, 2, 3, 4, &c, it is easy to see, that we 
have merely to set oft' in order from the beginning of the 
scale the several lengths, which shall be equal respectively 



52 PLANE TRIGONOMETRY. 

to the logarithms of these numbers, when ab is equal to the 
logarithm of 100. The logarithm of 100, taken from a com- 
mon table of logarithms but extending to three places only 
of decimals, is 2.000, or, omitting the separatrix, 2000. Un- 
der the same circumstances the logarithm of 2 is 301 , and the 
logarithm of 3 is 477, &c. Let it be supposed, that the line 
ab is divided into 2000 equal parts ; then, as ab has been 
made to represent the logarithm of 100, 301 of these parts 
will represent the logarithm of 2. To determine therefore 
the place of the number 2 on the scale, we set 301 of the 
parts, into which ab is supposed to be divided, from 1 to 2 ; 
to determine the place of 3 we set off in like manner 477 of 
the same parts, and so of the rest. Proceeding in this man- 
ner we determine the primary divisions of the scale, and to 
obtain the intermediate divisions we set off in a similar 
manner the logarithms of the intermediate numbers. Thus, 
when the logarithm of 100 is 2000, the logarithm of 1.1 is 41, 
the logarithm of 1.2 is 79, &c. These numbers being setoff 
in order from the beginning of the scale will divide the first 
primary division into 10 parts. In like manner the remain- 
ing primary divisions may be further divided. 

86. The logarithmic scale, the construction of which 
has now been explained, contains the logarithms of numbers 
from one to a hundred only. The decimal part of the loga- 
rithm of any number, it will be recollected, is the same as 
that of the number, when multiplied or divided by 10 or 100, 
&c. We may then consider the point under the 1 at the be- 
ginning of the scale to represent the logarithm of 1, or 10, 
or 100j or ^ or^, &c. Hence the same scale may easily be 
made to answer for all numbers whatever. 

87. From what ha$ been said the logarithm of any 
given number may with facility be taken from the scale. 

Let it be required to take from the scale the logarithms 



PLANE TRIGONOMETRY. 53 

of the following numbers, viz. 75, 43.5, 365, 2450, 10844. 

88. Multiplication, division, Sec. are performed by the 
line of numbers on the same principles as by common loga- 
rithms. 

Multiplication. 

1. Let it be required to multiply 6 by 8. Since addition 
in logarithms takes the place of multiplication, the extent 
from 1 to 6 added to the extent from 1 to 8 will be equal to 
the extent from 1 to 48 the product. 

To multiply then by the logarithmic line, we take off 
with the compasses that length of line which represents the 
logarithm of one of the factors, and apply this so as to ex- 
tend forward from the end of that which represents the log- 
arithm of the other factor. The sum of the two will reach to 
the end of the line, which represents the logarithm of the 
product. 

2. It is required to multiply 9 by 7, 18 by 5, 44 by 63, 
120 by 75. 

Division . 

It is required to divide 56 by 7, 120 by 12, 400 by 50, 
2880 by 320. 

Proportions . 

1. To find the fourth term of the proportion, in which 2, 
4 and 8 are the first three terms. 

2. To find the fourth term of the proportion, in which 15, 
75 and 40 are the first three terms. 

3. Let 4324, 540.5 and 2560 be the three first terms of a 
proportion ; required the fourth term. 

89. We proceed to construct a line of logarithmic sines, 
that shall correspond with the line of numbers. For this 
purpose let the line cd, intended for the scale, be taken 
(fig. 36) equal to ab. In order that this scale may corres- 

6 



54 PLANE TRIGONOMETRY 

pond with the line of numbers, it will be obvious, that the 
difference between the extreme indices on both should be 
the same. On the line of numbers the difference between 
the extreme indices is 2. The logarithmic sine of 0° 34' 
22" 41'", is 8.000, and that of 90° is 10.000 the difference 
of the indices being 2. If then the point c at the beginning 
of the scale be marked for the place of 0° 34' 22" 41'" the 
point d at the end of the scale will be the place for 90°. In 
order to find the intervening divisions we suppose the line 
cdy in the same manner as has been supposed with respect 
to ab, to be divided into 2000 equal parts. Then, since the 
difference between the logarithmic sine of 0° 34' 22" 41'" 
and the logarithmic sine of 1° is 241, the extent from c to 1 
equal to 241 of the parts, into which cd is supposed to be di- 
vided, will give the point of division for the sine of 1° ; in 
like manner the extent from 1 to 2, equal to 542 of the same 
parts, will give the point of division for 2°, and so of the 
rest. The scale may also be constructed by marking the 
point dfor the sine of 90°, and setting the arithmetical com- 
plement of each degree &c. backward from d toward c 

90. The logarithmic tangent of 0° 34' 22" 35'" is 8.000, 
and that of 45° is 10.000. To form therefore a line of loga- 
rithmic tangents corresponding to the line of numbers, let 
the point e, on the line ef taken equal to ab, be marked for 
the tangent of 0° 34' 22" 35'", and the point /for the tan- 
gent of 45° ; then, ef being supposed also to be divided into 
2000 equal parts, the intervening points of division may be 
marked as before. 

For the points of division above 45° the scale should ex- 

R _ cot 
tend much further to the right; but (art. 57) 7^"~— ~R~' 

that is, in logarithms R — tang = cot — R . The logarith- 
mic tangent therefore of an arc below 45° is as much less 



PLANE TRIGONOMETRY. 55 

than that of 45°, as the logarithmic cotangent of this arc is 
greater than that of 45°. Instead then of extending the 
scale further to the right, the numbering after reaching 45° 
may be continued back from right to left, and the same 
point of division be made to answer for an arc and for its 
complement. 

91. We shall now show the use, which may be made of 
the logarithmic lines in the solution of triangles. 

Ex. 1. In the triangle ABC (fig, 30), given the side AB 
40 yds, the side AC 90 yds, and the angle C 20°, to find by 
means of the logarithmic lines the angle B. 

By art. 66, AB : AC : : sin C : sin B ; wherefore in loga- 
rithms AC — AB = sinB — sin C ; hence to find the angle 
at B, we take in the compasses from the line of numbers the 
extent from 40 to 90 ; this extent will reach from 20° on 
the line of sines to 50° 20' the answer. 

2. Let the angle C equal 55°, the angle B 32°, and the 
side AB 100 feet to find AC. 

3. Given AB 53, BC 13, and the angle B, 120°, to find 
AC. 

92. The scale fig. 36, the construction of which has now 
been explained, is commonly called Gunter's scale from the 
name of its inventor. It furnishes a ready method of solv- 
ing triangles in those cases, where great accuracy is not re- 
quired. The mode of solution differs from that by the com- 
mon tables of logarithms in this respect only, that the loga- 
ithmic values of the sides and angles are expressed by lines 
instead of numbers. 

SECTION XV. 

Of Trigonometrical Analysis. 

93. In the preceding sections we have treated of sines, 
cosines, tangents, &c. merely in their relation to the calcula- 



56 PLANE TRIGONOMETRY. 

tion of the sides and angles of triangles. The use of these 
lines may however be extended to other objects. In conduc- 
ing the investigations of physical astronomy in particular 
important assistance is derived from them; because as it is 
only the angular positions of the heavenly bodies, which we 
observe, all disquisitions concerning the motions, orbits, &c. 
of these bodies must necessarily involve the sines, cosines, 
tangents, &c. of angles. 

94. In the solution of plane triangles the arcs, which fall 
under consideration, never exceed a semicircumference ; and 
we have occasion to consider merely the numerical values of 
the sines, cosines, tangents, &c. In the more extended ap- 
plications of these lines, however, the arcs, which occur, as- 
sume every possible magnitude, and the sines, cosines, &c. 
being regarded as functions of their corresponding arcs, that 
is, as analytic expressions, the values of which depending 
entirely upon those of the arcs may be deduced from them 
by certain arithmetical operations, we shall have occasion 
to consider not only the changes that take place in the nu- 
merical values of the different trigonometrical lines, but also 
the changes that occur in their positions as the arcs to which 
they belong vary in magnitude. 

95. To understand clearly the nature of these changes 
let us attend to the variations, which take place in the nu- 
merical values, and in the positions of the sines, cosines.tan- 
gents, kc. of an arc, while it increases from to an entire 
circumference. 

In the circle ABDE (fig. 34) let the diameters AD, BE 
be drawn perpendicular to each other ; the circle will then 
be divided into quadrants. To trace the changes in question, 
let the radius CF, at first coinciding with CA, be supposed 
to revolve about the point C, as upon a pivot, and departing 
from the position CA, so that the arc AF shall have succes- 



PLANE TRIGONOMETRY. 57 

sively all magnitudes whatever fom to an entire circumfer- 
ence, and the angle ACF shall increase from to four 
right angles. In the first quadrant, when the radius CF 
coincides with CA, so that the arc AF is 0, the sine GF, it 
is evident, is also 0, while the cosine CG is equal to radius 
or unity. When the radius CF moves off from CA, the sine 
GF increases as the point F advances toward B, until, when 
the point F has arrived at B, the sine GF coincides with CB, 
and becomes equal to radius or 1. At the point B the arc 
AF is equal to a quadrant, and the angle ACF to a right 
angle. Under the same circumstances the cosine CG 
constantly decreases, until when the point F coincides 
with B, the cosine becomes equal to 0. In the second 
quadrant as the point F' moves on from B to D, the 
sine G'F' decreases, and the cosine CG' increases, until, 
when the point F' coincides with D, and the arc AF' be- 
comes equal to a semicircumference, the sine G' F' is 0, and 
the cosine C G' is equal to radius or 1. 

It may be remarked in passing, that the lines G'F', CG" are 
respectively the sine and cosine of the arc DF', and also of its 
supplement AF'; whence the absolute magnitude of the sine 
and cosine of an obtuse angle is the same with that of its 
supplement . 

In the third quadrant, as the point F" moves on from D 
to E, the sine G" F" increases, and the cosine CG" decreases, 
until at the point E, where the arc AF" becomes equal to 
three fourths of a circumference the sine is equal to 1, and 
the cosine to 0. From E to A the sine G'"F"' decreases 
and the cosine CG"' increases, until at A, when the whole 
revolution has been completed, and the arc AF'" become^ 
equal to an entire circumference, the sine is reduced to 0, and 
the cosine becomes equal to 1. 
6* 



£8 PLANE TRIGONOMETRY. 

In the first and second quadrants the sines, being all sit- 
uated on the upper side of the diameter AD, from which 
they are measured, are considered positive] in the third and 
fourth quadrants falling below the diameter AD, and assu- 
ming in consequence a direction contrary to that, which 
they had before, they are considered negative. In the first 
and fourth quadrants the cosines, which are all situated on 
the left of the point C, to which they are referred, are con- 
sidered positive ; in the second and third quadrants falling on 
the right of C, and assuming a direction contrary to their 
former direction, they are considered negative 

96. Observing in like manner the changes, which occur 
in the magnitude of the tangents through the course, which 
lias been described, we find, that from A to B the tangents 
increase continually, as the arc AF increases ; at the point 
B where the arc AF becomes equal to a quadrant, the se- 
cant CM coinciding with CB is parallel to the tangent AM, 
and therefore no longer meets it, so that the arc AB has not 
properly speaking a trigonometrical tangent. We say, in- 
deed, that the tangent of AB or 90° is infinite ; by this ex- 
pression, however, we mean, that if the difference between 
an arc and 90° is indefinitely small, the tangent of this arc 
will be indefinitely great, that is, greater than any assign- 
able quantity. From B to D the tangents decrease, until at 
D the tangent is 0; from this point they increase, until at 
E the tangent is again infinite ; from E they decrease, and 
the tangent becomes at A. 

With respect to the changes, which occur in the positions 
df the tangents, from A to B the radius CF produced meets 
the indefinite tangent at A above the diameter AD; in the 
first quadrant therefore we consider the tangents positive. 
From the point B to D, the radius F'C produced no longer 
meets the indefinite tangent at A above the diameter AD ;? 



PLANE TRIGONOMETRY. 59 

but below it ; in the second quadrant then the tangents are 
negative. Tracing in like manner the changes of position, 
which take place with respect to the tangents in the remain- 
ing quadrants, we find, that in the third quadrant they again 
become positive, and in the fourth negative. 

97. We have traced the changes, which occur with re- 
spect to the sines, cosines, &c. from the departure of the 
point F from A until its return to this point again. We 
may now suppose that at A it commences a second revolu- 
tion, and regarding as one arc the whole course passed over 
by the point F from the commencement of its motion, we 
shall have arcs, that exceed a circumferance, and which have 
the same sines, cosines, tangents &c. as those described in 
the first revolution. 

93. The formulas for the sine and cosine of the sum, and 
the difference of two arcs obtained by a geometrical demon- 
stration art. 35, 36 are true whatever the magnitude of 
these arcs. Representing the arcs by the letters a and b 
respectively, those formulas, radius being considered as uni- 
ty, may now be expressed in algebraic language as follows 

1. sin (a-\-b =sin a cos 6 -f- sin b cos a 

2. sin (a — 6) = sin a cos b — sin b cos a 

3. cos (a -j- 6) = cos a cos b — sin a sin b 

4. cos (a — b) = cos a cos b -f- sin a sin b 

99. The four preceding formulas form the basis of trigo- 
nometrical analysis. We shall now employ them in the pre- 
paration of various other formulas of the greatest utility in 
the more extended application of the trigonometrical lines. 
Before making use of them, however, for this purpose, it 
may be well, in order to see the precise agreement between 
the analytic and geometrical mode of considering the subject, 
to show from these expressions the variations, both in sign 
and magnitude, of the sines and cosines through the first 
circumference. 



60 PLANE TRIGONOMETRY. 

Let p denote a semicircumference ; then the cosine of 
\ p is 0, and its sine is 1 ; hence substituting | p for a in the 
preceding formulas, we have 

sin (| p ±b)= cos 6, cos (|- p ± 6) = =F sin 6. 
To show the absolute value of these expressions, let the 
arc BF' (fig. 34) = 6; since the arcAB=|p, the arc 
AF' =r (| p + 6) ; then G'F' the sine of AF', and also of 
DF', is of consequence the cosine of BF'; hence sin (|p 
-]-&)== cos 6; CG' moreover the cosine of AF' is the sine 
of BF' ; hence cos (| p -f b) = sin 6 . If we make BF = 6, 
then sin ( § p — b) = cos 6, and cos (± p — 6) = sin 6. 

With respect to the sign — , which affects the absolute 
value of cos (| p -\- b), it signifies, that if we regard as posi- 
tive the sine and cosine of an arc less than a quadrant, the 
cosine of an arc greater than a quadrant, but less than a semi- 
circumference will be negative, while its sine is positive. 

If we make b = a quadrant, then sin (§ p db 6) = 0, and 
cos (i p db b) = =F 1 • Again let a = p, then sin (p ± 6)= 
q± sin 6, and cos (p zb 6) = — cos b ; if we make b = | p, 
then sin | p = — 1 , and cos | p = 0. Lastly let a = f p, then 
sin (| p =b &)= — cos 6, and cos (§ p ± 6) = ± sin 6. The 
absolute values of these last expressions may also be eas- 
ily verified ; the signs show that every arc comprehended 
between p and | p has its sine and cosine negative, while 
an arc comprehended between |p and 2p has its sine neg- 
ative, but its cosine positive. 

Expressions for the sines mid cosines of multiple arcs. 
100. Putting in the expression for sin (a -|- b) no. 1 art. 
98, b = a, 2 a, 3 a, &c, we shall have 
sin 2a = 2 sin a cos a 
sin 3o = sin a cos 2 a -j- sin 2 o cos a 
sin 4 a = sin a cos 3 a -}* s ^ 3 a cos a 
sin 5o = &e 



PLANE TRIGONOMETRY. CI 

Making the same substitutions iu the expression for cos 
(a -J- b) no. 3 art. 98, we have , 

cos 2 a = cos 2 a — sin 2 a 

cos 3 a = cos a cos 2 a — sin a sin 2 a 

cos 4o = cos a cos 3 a — sin a sin 3 a 

cos 5 a = &c. 
By the expressions sin 2 a, cos 2 a, &c, we understand the 
square of the sine of a, square of the sosine of a, &c. 

101. To find expressions for sin 3 a, sin 4 a, cos 3 a, cos 4 
a &c in terms of the simple arc, we substitute in the formulas 
just obtained the values of sin 2 a , cos 2 a &c. Thus we 
have sin 3 a = 3 sin a cos 2 a — sin 3 a 

sin 4 a = 4 sin a cos 3 a — 4 sin 3 a cos a 

sin 5 a= &c. 

cos 3o= cos 3 a — 3 sin 2 a cos a 

cos 4 a = cos 4 a — 6 sin 2 a cos 2 a -(- sin 4 a 

cos 5 a = &c. 

Expressions for the sines and cosines of half a given arc 

102. In the expresssion for cos 2 a art. 100, making a = 
| a, we have cos 2 \a — sin 2 |a== cos a (1) 
but art. 30 cos 2 \ a -f- sin 2 1 a = 1 (2) 
whence by subtraction 2 sin 2 -| a = 1 — cos a 
wherefore sin | a == (| — ^ cos a) a 

Adding next the equations 1 and 2, and proceeding in 
like manner, Ave have cos \ a= (| -[~i cos a ) « 

Expressions for the products of sines and cosines. 

103. Adding the equations 1 and 2 art. 98, we have 

sin (a -f- 6) -f- sin (^a — 6) = 2 sin a cos 6 
whence sin a cos 6 = \ sin (a -f- 6) -{- J sin {a — b) (1) 

Subtracting the second of the same equations from the 
first and reducing, we have 

sin 6 cos a = J sin (a -+ 6) — | sin (a — b) (2) 



62 PLANE TRIGONOMETRY. 

In like manner the equations, 3 and 4 art. 98 give 

cos a cos 6 = \ cos (a -f- b) -J- i cos (a — 6) (3) 

sin a sin 6 = | cos (a — 6 ) — \ cos (a -{- 6) (4) 

Expressions for the powers of sines and cosines. 

104. 1. Let b = a in nos 3 and 4 of the preceding article, 
and we have 

sin 2 a = \ — | cos 2 a 
cos 2 a == | + | cos 2 a 
2. Multiplying the first of the preceding equations by 
sin a, or to avoid fractions by 4 sin a, we have 

4 sin 3 a = 2 sin a — 2 sin a cos 2 a 
but art. 103 sin (a + b) — sin (a — b) = 2 sin 6 cos a 
in this last equation let a = 2 a and b==a, then 

sin 3 a — sin a = 2 sin a cos 2 a 
wherefore by substitution and reduction 
3 sin a — sin 3 a 



4 

and by a process altogether similar, we obtain 

3 cos a -f- cos 3 a 
cos * a = 4 

3. In like manner, we find 

cos 4 a — 4 cos 2 a -f- 3 

sin 4 a = 

8 

cos 4 a + 4 cos 2 a + 3 
cos 4 a = g 

Expressions for the tangents of arcs in terms of the tangents. 

sin ( a -f-6) 
105. By art. 48, we have tan (a + 6) = J^-^ZJTft), 

sin a cos 6 -f- sin b cos a 

whence by substitution tan (a -f- 6) = 

cos a cos 6 — sin a sin b 

dividing the numerator and denominator of the second 



PLANE TRIGONOMETRY. 63 

member of this equation by cos a cos 6, we have for the Enu- 
merator 

sin a cos 6 -4- sin t cos a sin a , sin 6 

_ — = -p _ f an a i tan 6 

cos a cos 6 cos a cost ' 

and for the denominator 

cos a cos 6 — sin a sin 6 . sin a- sin 6 A 

= = 1 — X r= 1 — tan a tan b 

cos a cos b cos a cos 6 

whence by substitution and reduction, we obtain 

, , tan a -4- tan b 
tan (a + 6 )= ^ (1) 

1 — tan a tan b 

In like manner we obtain tan (a — b) =— — (2) 

v J l+tanatan6 v ' 

By art. 57, we have cot (a -f- 6) = — — — _ ? w hence 

by substitution 1 —tan a tan b 

cot (a + l)= — (3) 

tan a -\- tan b 

1 -j- tan a tan 6 

So also cot (a — b) = (4) 

tan a — tan b 

Expressions for the tangents and cotangents of multiple arcs in 
terms of the simple arcs. 

106. Putting successively b = a, 2 a, 3 a,&c, in nos 1 and 
3 of the preceeding article and substituting for factors of 

multiple arcs, we have 

2 tan a 

tan 2 a = - 



tan 3 a = 



1 — tan 2 a 
3 tan a — tan 3 a 



1 — 3 tan 2 a 
4 tan a — 4 tan 3 a 



tan 4 a = 

1 — 6 tan 2 a -f- tan 4 a 

tan 5 a = &c. 

The expressions for the cotangents, will it is easy to see,, 
be merely an inversion of the preceding. 



64 PLANE TRIGONOMETRY. 

Expressions for the tangents and cotangents of half an arc. 

107. By art. 48 we have tan | a = — [ -\— > whence by sufc- 
J 2 cos i a 

stitution art. 102 (1 — cos a ) $ 

V~2 ( 1— cos a) * 
tan |a = -— = v '— 

(1 -fcosajs (l-}-cosa)2 

multiplying numerator and denominator hy (1 -\- cos a ) & 

(1 — cos 2 a) 2 sin a 

tan lo= — — , — ==— ■ 

1 -j- cos a 1-fcos a 

The expression for the cotangent | a, it is evident, will 

be the inverse of that for the tangent | a. 

Expressions for secants and cosecants. 
108. By art. 48 we have sec (a -J- b) — 



cos (a -{- 6) 

whence by substitution 1 

sec (a-)- 6): 



cos a cos 6 — sin a sin 6 
dividing both numerator and denominator by cos a cos b and 
substituting we have 

, x sec a sec 6 

sec (a -f-6) = — 

- 1 — tan a tan 6 

: - sec a sec b 

S9 also sec (a - 6) = 2 + tan a tan 6 ' 

And by a process altogether similar we obtain 

sec a sec b 

cosec (a ± 6) = 

tan a ± tan 6 

Making 6= a in the expressions for sec and cosec a -{- 6, 
we have 

sec 2 a 

sec 2 a = — r* 

1 — tan * a 

sen 2 a 
cosec 2 a =WJ^~ 



PLANE TRIGONOMETRY. 65 

109. From what has been done, it will be perceived, 
that an almost infinite variety of trigonometrical formulas 
may be made adapted to the purposes, for which they may be 
required ; we shall close the subject, however, with the in- 
vestigation of formulas for finding the angle of a triangle 
when the sides are given, and also for determining the area 
of a triangle from different data. 

110. In the triangle ABC (fig. 33), let fall from one of 
the angles, B for example, upon the side AC the perpendicu- 
lar BD ; then(Geom. 191, 192) AB 2 = AC 2 + BC 2 =F2 AC 
X DC, the sign — being used when the perpendicular falls 
within the triangle, and the sign -J- when it falls without ; 
but in the right angled triangle BDC, radius being 1, DC = 
BC X cos C, wherefore AB 2 =AC 2 + BC 2 — 2 AC X BC 
X cos C, the sign — being sufficient for the term 2 AC &c. 
because when the angle C is obtuse its cosine is negative, 
and consequently changes — into -f- as is required by the 
geometrical construction. Hence, employing the letters a, 
by c, to represent the sides BC, AC, AB, and deducing the 

value of cos C, we have a 2 + 6 2 c 2 

cos C = — > 

2 a b 

but this expression not being well adapted to calculation by 

logarithms, another is to be sought. 

The formula sin 2 a = \ -^ \ cos 2 a gives by substituting 

\ C for a, and deducing the value of cos C, 

cosC=l — 2sin 2 |C 

comparing the two values of cos C, and deducing that of 

2 sin 2 \ C, we have 

c 2 — a t ^b 2 c 2 — a 2 — b * 4- 2 a h 

2sin ! 4C=l+ — = ■ ! 

2 ab Zab 

_ c> — (q^6) 2 _ (c + «— V)(c — aJ r b ) 
— 2a6 "" 2a6 

7 



66 PLANE TRlGONpMETRr. 

whence 

( c+g-6)(c^a + 6 ) |( c + a-&)|(c--a-f 6) 

sm 2 iC= — f = 

2 4 a6 aft 

but | (c -j- a : — 6) = J (c -f- a + 6) — 6, and i (c — a -f- 6)=s 

i (c -f- a + 6) — a, whence representing c -f- a + 6 by s and 

reducing, we have 

sin i c = \ ^ / 

deducing next the value of cos C from the formula cos 2 a = 
£ -j- § cos ^ a > we °°tain hy a process altogether similar 



.c^iil-f 



cos ^ 

ab 

Expressions for the surfaces of triangles. 

1. Given two angles and the included side to find the 
surface of the triangle. 

Let AC (fig. 33) be the given side A and C the given 
angles ; employing the same notation as above, we have 
6 sin C b sin C 
sin B sin (A 4- C) 
the sine of an angle being the same with that of its supple- 
ment. Calling d the perpendicular let fall upon the given 
side, we have d = c sin A, whence 
6 sin A sin C 
d = sin(A + Cy 
multiplying this expression by \ b in order to obtain the sur- 
face sought, and denoting this surface by S, we have 
6 2 sin AsinC 
2sin(A + C) 
Substituting next for sin (A + C) its value art; 98', and di- 
viding both numerator and denominator by sin A sin C, we 

have finally & 2 

S sS 



2 (cot A -f cot C) 



PLANE TRIGONOMETRY 67 

2. Given two sides and the included angle to find the 
surface of the triangle. 

Let a and 6 be the given sides (fig. 33), C the given an- 
gle, we have d = a sin C ; whence multiplying the perpen- 
dicular by | b we have 

S = \ ab sin C 

3. Given the three sides to find the surface of the trian- 
gle. 

In the triangle ABC (fig. 33) we have DC = a cos C, 
designating BD by d as before, we have by the known 
property of a rjght angled triangle. 

d 2 =a 2 — a 2 cos 2 C 
Substituting for cos C its value in art. 110 we have 
/ a * -f- b 2 — c 2 \ 2 

d2 =« 2 -v — v, — ) 

whence 4b 2 d 2 = 4a 2 b 2 — (a 2 + 6 2 — c 2 ) 2 

but the two terms of the right hand member being the dif- 
ference between two squares, we have 
4 b 2 d 2 = (2 ab -f a 2 -f b 2 — c 2 ) (2 ab — a 2 — b 2 + c 2 ) 

= ((a + 6) 2 -c 2 )(c 2 -(o-&) 2 ) 
the two factors of this last being each also the difference 
between two squares, we have 

4i 2 d 2 =(a + 6 + c) (a + b — c) (c +a — b) (c + 6 — a) 
adding and subtracting c, b and a in the second third and 
last factor respectively, and designating a -j- b -j- c by 2 s, 
we have 

4 6 2 d 2 = 2 s (2 s — 2 c) (2 s — 2 6) (2 s — 2 a) 
= 16s (s — c) (s— 6) (s — a) 
dividing both sides by 16 and extracting the square root 

Hence, to find the surface of a triangle by means of the 
three sides, we must subtract successively each side from the half 



68 PLANE TRIGONOMETRY. 

sum, multiply the half sum and the three remainders continually 
together, and lake the square root of this product. 

MISCELLANEOUS PROBLEMS. 

1. A flag-staff is placed upon a wall 40 feet long in such 
a situation, that a line of 34 feet in length will reach from 
its top to one end of the wall and a line of 2d feet from its 
top to the other end. What is the height of the flag-staff, 
and its distance from the ends of the wall ? 

2. From the top of a tower, whose height is known, was 
ohserved the angular distance between two objects situated 
in the same horizontal plane with its base, and also the angles 
of depression of each of them. To determine from these 
data the distance between the two objects, the objects not 
being in the same horizontal plane. 

3. Let A and C be two stations on a sloping ground dis- 
tant 410 yards, an object on the top of a hill ; the following 
angles being found by observation, viz. OCA 79° 29', OAC 
63° 11', and the angles of elevation at A and C 6° 36' and 
5° -2-2' respectively ; it is required to calculate the height 
and distance of the object from each station. 

4. At the same instant two observers take the elevation 
of a balloon D at two stations A and B, and also the angles 
DAB, DBA ; the distance AB is m yards and the elevation 
of B above A is n° ; to determine from these data the eleva- 
tion of the balloon and its distance from the lower station A. 

5. The distances between three points A, B and C (fig.48) 
are given, and also at a point D in the line BA produced the 
angle ADC is found by observation to be ?i°, to determine 
the distance from the point D to each of the points A, B 
and C. 

6. From the top and bottom of a building, whose height 
is known, the elevation of another building on the same hori- 



PLANE TRIGONOMETRY. C9 

zontal plane is observed, to find that point in the plane,which 
is equidistant from the top of each. 

7. How must three trees A, B, C, be planted, so that the 
angle at A may be double the angle at B, and the angle at 
B double the angle at C and a line of 400 yards may just go 
round them? 

8. The distances between three points A, B, C (fig. 43) 
being known,it is required to find the position of the point D 
with reference to these points, the angles ADB, ADC being 
found by observation equal to m° and n° respectively. 

Let the given distances AB, AC, BC be represented 
by the letters a, A, c, respectively ; in the triangle ADB let 
the angle ABD be represented by #, and in the triangle 
ADC the angle ACD by y; the triangles ADB ADC give 
the equation. 

a sin n sin x = b sin m sin y 
but in the quadrilateral ABDC we have 

y = 4 right angles — ADB — ADC — BAG — x. 
"but since the angle BAC is known when the three sides of 
the triangle ABC are given, the first four terms of the right 
hand member of this last equation are known ; representing 
these by d for the sake of conciseness, and substituting for y 
its value in the first equation, we have 

a sin n sin x = b sin m sin (d — x) = b sin m (sin d cos x — 
sin x cos d ), dividing both sides by sin x we have 

a sin n-=b sin m (sin d cot x — cos d) 

a sin n -\- b sin m cos d 

whence c °t x — ; — : : — j- 

b sin m sin a 

9. A person travelling along a road BA (figure 49) 
observes at A the elevation of a tower CD, the perpendicu- 
lar distance of which BD from the road is known. At the 
same time he also observes the angular distance CAB of 
the top of the tower from an object B in the road. To deter- 

' mine from these data the height of the tower. 



70 PLANE TRIGONOMETRY. 

In the triangle CAD we have R : AC : : sin CAD : CD 
and in the triangle CAB R : AC : : sin CAB : CB 

whence sin CAD : sin CAB : : CD : CB 

but iia the triangle CBD, we have CD : CB : : sin CBD : R 

comparing this last with the preceding, we obtain R being 1 

. ^™^ sin CAD 

sin CBD = 

sin CAB 

the value of CBD being determined, the height CD is readi- 
ly found. 

10. At a station in a right line joining the bases of 
two mountain peaks I observed the angle of elevation of each 
of the peaks ; proceeding thence 100 yards further on level 
ground in a direction perpendicular to the line joining the 
bases of the peaks, I again observed their angles of eleva- 
tion. How may the distance between the two peaks be de- 
termined from these data. 

11. From a window A in the side of a tower exactly 50 
feet from the ground, the angle of depression of the bottom 
of a may-pole was found to be m°, it was also observed that 
the angle of elevation of the top of the pole at this station 
was equal to the complement of its angle of elevation at the 
bottom of the tower ; to determine from these data the 
height of the tower, (fig. 51). 

By the question the side AB and the angle BAD are given, 
and the angle EAC is the complement of ABC ; the angle 
ACD is equal to 90° — ABC, and the angle ACB to 90°— 
2 ABC. 

In the triangle ACB we have sin ACB : AB : : sin ABC : AC 
and in the triangle ACD sin ACD : AD : : sin ADC: AC 
whence AD sin ADC __ AB sin ABC 

sin (90°— ABC) sin (90°— 2 ABC) 

AD sin ADC AB sin ABC 

or = 

cos ABC cos 2 ABC 



PLANE TRIGONOMETRY. 7j 

from which we have art. 37, 48, tan 2 ABC == ^A D sin ADC 

AB 
AD being found by means of the triangle ABD, we have by 
the above formula the angle ABC, and this being found the 
height sought will readily be determined. 

12. Having at a certain unknown distance taken the angle 
of elevation DAB of a Steeple DE (fig. 44) I advanced to a 
point B 60 yards nearer on level ground and there observed 
the angle of elevation to be the complement of the former, 
advancing still nearer 20 yards to another point C, the an- 
gle of elevation there appeared to be just double of the first. 
To determine the altitude of the steeple. 

Putting DAB =#,we have by the question DBC =90° — ar, 
and DCE = 2x ; whence BDC = 3x — 90°, and ADB = 
90° — 2x 

in the triangle ADB we have sin (90° — 2x) : 60 : : sin x : DB 
and in the triangle BDC sin (3 x — 90°) : 20 : : sin 2 x : DB 
whence 20 sin 2x 60 sin x 

sin (3ar~90°) = sin (90°— 2x) (0 

but the triangle ACD being isosceles, we have DC = 80 
yards, whence in the triangle BDC we have art. 69 
100 : 60 : : sin (90° — x) + sin (3 x — 90°) : : sin (90° — x) 
— sin (3 x — 90°), from which we obtain sin (90 Q — x) = 
4 sin (3 x — 90°) whence by substitution in equation (1) 
20 sin 2 x 60 sin x 

IlhT(90° — x) === sin (90° — 2~x) 
substituting next for sin 2 x its value in terms of x, and for 
the sines of the angles in the denominators the cosines of 
their complements, we obtain finally cos 2 x= . 37500, and 
the angle 2 # or DCE being thus obtained the height DE 
will be easily found. 

13. A tower whose height is 60 feet is situated on the 
side of a hill, which rises due N. at an angle BDE of 20° ; 



72 PLANE TRIGONOMETRY. 

what length of ladder will be necessary to scale the tower, 
supposing the foot of the ladder to be placed at a distance 
BC of 30 feet in a direction S 40° W from the base of the 
tower, (fig. 52). 

The lines BD and BC being referred to a horizontal 
plane, we have by the question the horizontal angle CED = 
40° ; CD moreover being supposed perpendicular to DE, 
the triangle CDE will be right angled at D. 

From the triangle BCE we obtain radius being 1, BE = 
30 sin BCE, and from the triangles BCE, DCE, DE = 30 
sin 50° cos BCE. 
in the triangle BDE we have 

DE + BE : DE — BE : : sin 70° -f sin 20° : sin 70<* — 
sin 20° 

putting for the sake of conciseness the letters a and b for the 
last two terms of this proportion respectively, and substitu- 
ting for DE and BE the values found above, we have 

(a -f b ') sin BCE = (a — 6) sin 50° cos BCE 

(a — b) sin 50° 
whence tan BCE = 

(a + 6) 

and BCE being determined by this formula the length sought 
will be readily found. 



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